How do I check if a number is a palindrome

Question

How do I check if a number is a palindrome   Any language  Any algorithm   except the algorithm of making the number a string and then reversing the string

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Push each individual digit onto a stack  then pop them off   If it s the same forwards and back  it s a palindrome

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num   int raw input    list num   list str num   if list num   -1     list num      print  Its a palindrome  else      print  Its not a palindrom

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public boolean isPalindrome int x            if  isNegative x               return false           boolean isPalindrome   reverseNumber x     x   true   false          return isPalindrome             private boolean isNegative int x            if  x  lt  0              return true          return false             public int reverseNumber int x             int reverseNumber   0           while  x  gt  0                int remainder   x   10              reverseNumber   reverseNumber   10   remainder              x   x   10                     return reverseNumber

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public static void main String args          System out print  Enter a number          Scanner input   new Scanner System in       int num   input nextInt        int number   num      int reversenum   0      while  num    0                reversenum   reversenum   10          reversenum   reversenum   num   10          num   num   10             if  number    reversenum          System out println  The reverse number is     reversenum     nThen the number is palindrome         else         System out println  The reverse number is     reversenum     nThen the number is not palindrome

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let isPalindrome  n int       let l1   n ToString     gt  List ofSeq   gt  List rev    let rec isPalindromeInt l1 l2          match  l1 l2  with           h1  rest1 h2  rest2  - gt  if  h1   h2  then isPalindromeInt rest1 rest2 else false            - gt  true    isPalindromeInt l1  n ToString     gt  List ofSeq

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Not sure if this is the best answer  and I m new at programming   it s in java   import java util     public class Palin        public static void main String   args            Random randInt   new Random             Scanner kbd   new Scanner System in           int t   kbd nextInt        of test cases          String   arrPalin   new String t     array of inputs          String   nextPalin   new String t           for  int i   0  i  lt  t  i                  arrPalin i    String valueOf randInt nextInt 2147483646    1               System out println arrPalin i                       final long startTime   System nanoTime             for  int i   0  i  lt  t  i                  nextPalin i     unmatcher incrementer switcher match arrPalin i                           final long duration   System nanoTime   - startTime           for  int i   0  i  lt  t  i                  System out println nextPalin i                       System out println duration               public static String match String N            int length   N length               Initialize a string with length of N         char   chars   new char length           Arrays fill chars   0             int count   1           for  int i   0  i  lt  length  i                  if   i 2     0      at i   even                  if  i    0                        chars i    N charAt i                     else                     chars i    N charAt i 2                 else   at i   odd                 chars i    N charAt length - count                   count                       return String valueOf chars              public static String switcher String N            int length   N length            char   chars   new char length           Arrays fill chars   0             for  int i   0  i  lt  length  i                  if  i    0                    if   i   2     0                        chars i    N charAt i                     else if   i   2     0                        chars i    N charAt i - 1                                               if  i    0                    chars 0    N charAt 0                                   return String valueOf chars              public static String incrementer String N            int length   N length            char   chars   new char length           Arrays fill chars   0             char   newN   N toCharArray             String returnVal           int numOne  numTwo           if   length   2     0                numOne   N charAt length-1               numTwo   N charAt length-2               newN length-1     char  numOne 1               newN length-2     char  numTwo 1               returnVal   String valueOf newN             else               numOne   N charAt length-1               newN length-1     char  numOne 1               returnVal   String valueOf newN                     return returnVal             public static String unmatcher String N            int length   N length            char   chars   new char length           Arrays fill chars   0            char   newN   N toCharArray             for  int i   0  i  lt  length  i                      if    i   2     0   amp  amp   i    0        for i  gt  0  even                     newN i   2    N charAt i                     else if   i   2     0  amp  amp   i    0        for i   0                     newN 0    N charAt 0                                           for  int i    length 2   i  lt  length  i                  newN i    newN Math abs i -  length - 1                        return String valueOf newN               So for this code  input a number  I did random numbers of how many I want    it will convert  for example the input is 172345 to 157423  Then it will change it to 117722  then if even make it 117733  if odd do the same for the only the last digit  Then make it 173371  It s not specifically finding a palindrome  but it s finding the next highest palindrome number

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int is palindrome unsigned long orig        unsigned long reversed   0  n   orig       while  n  gt  0                reversed   reversed   10   n   10          n    10             return orig    reversed

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def ReverseNumber n  partial 0       if n    0          return partial     return ReverseNumber n    10  partial   10   n   10   trial   123454321 if ReverseNumber trial     trial      print  It s a Palindrome      Works for integers only  It s unclear from the problem statement if floating point numbers or leading zeros need to be considered

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I always use this python solution due to its compactness   def isPalindrome number       return int str number    -1    number

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Here is an Scheme version that constructs a function that will work against any base  It has a redundancy check  return false quickly if the number is a multiple of the base  ends in 0   And it doesn t rebuild the entire reversed number  only half  That s all we need    define make-palindrome-tester     lambda  base        lambda  n          cond              0  modulo n base    f            else            letrec                 Q  lambda  h t                       cond                          lt  h t   f                            h t   t                         else                         let                               h2  quotient h base                                m   - h    h2 base                               cond                                h2 t   t                              else                              Q h2       base t  m                        Q n 0

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This is one of the Project Euler problems   When I solved it in Haskell I did exactly what you suggest  convert the number to a String   It s then trivial to check that the string is a pallindrome   If it performs well enough  then why bother making it more complex   Being a pallindrome is a lexical property rather than a mathematical one

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here s a f  version   let reverseNumber n       let rec loop acc   function      0 - gt  acc      x - gt  loop  acc   10   x   10   x 10          loop 0 n  let isPalindrome   function       x  when x   reverseNumber x - gt  true         - gt  false

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def palindrome n       d          while  n  gt  0           d append n   10          n     10     for i in range len d  2           if  d i     d - i 1                 return  Fail       return  Pass

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For any given number   n   num  rev   0  while  num  gt  0        dig   num   10      rev   rev   10   dig      num   num   10      If n    rev then num is a palindrome   cout  lt  lt   Number    lt  lt   n    rev    IS     IS NOT    lt  lt    a palindrome   lt  lt  endl

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This is my Java code  Basically comparing the the first and last value of the string and next inner value pair and so forth         Palindrome number       String sNumber    12321       int l   sNumber length       getting the length of sNumber  In this case its 5     boolean flag   true      for  int i   0  i  lt   l    i            if  sNumber charAt i     sNumber charAt  l--  -1       comparing the first and the last values of the string             System out println sNumber    is not a palindrome number                flag   false              break                      l--     to reducing the length value by 1            if  flag            System out println sNumber    is a palindrome number

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Push each individual digit onto a stack  then pop them off   If it s the same forwards and back  it s a palindrome

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Just for fun  this one also works   a   num  b   0  if  a   10    0    return a    0  do     b   10   b   a   10    if  a    b      return true    a   a   10    while  a  gt  b   return a    b

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Check this solution in java     private static boolean ispalidrome long n            return getrev n  0L     n             private static long getrev long n  long initvalue            if  n  lt   0                return initvalue                    initvalue    initvalue   10     n   10           return getrev n   10  initvalue

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public class Numbers       public static void main int givenNum               int n  givenNum        int rev 0          while n gt 0                       To extract the last digit           int digit n 10               To store it in reverse           rev  rev 10  digit               To throw the last digit           n n 10                   To check if a number is palindrome or not       if rev  givenNum                    System out println givenNum  is a palindrome                   else                  System out pritnln givenNum  is not a palindrome

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int reverse int num        assert num  gt   0        for non-negative integers only      int rev   0      while  num    0                rev   rev   10   num   10          num    10            return rev      This seemed to work too  but did you consider the possibility that the reversed number might overflow  If it overflows  the behavior is language specific  For Java the number wraps around on overflow  but in C C   its behavior is undefined   Yuck   It turns out that comparing from the two ends is easier  First  compare the first and last digit  If they are not the same  it must not be a palindrome  If they are the same  chop off one digit from both ends and continue until you have no digits left  which you conclude that it must be a palindrome   Now  getting and chopping the last digit is easy  However  getting and chopping the first digit in a generic way requires some thought  The solution below takes care of it   int isIntPalindrome int x        if  x  lt  0      return 0      int div   1      while  x   div  gt   10                div    10             while  x    0                int l   x   div          int r   x   10          if  l    r              return 0          x    x   div    10          div    100            return 1

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For any given number   n   num  rev   0  while  num  gt  0        dig   num   10      rev   rev   10   dig      num   num   10      If n    rev then num is a palindrome   cout  lt  lt   Number    lt  lt   n    rev    IS     IS NOT    lt  lt    a palindrome   lt  lt  endl

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Personally I do it this way  and there s no overlap  the code stops checking for matching characters in the correct spot whether the string has an even or odd length  Some of the other methods posted above will attempt to match one extra time when it doesn t need to    If we use length 2  it will still work but it will do one additional check that it doesn t need to do  For instance   pop  is 3 in length  3 2   1 5  so it will stop checking when i   2 since 1 lt 1 5 it will check when i   1 as well  but we need it to stop at 0  not one  The first  p  is in position 0  and it will check itself against length-1-0 current position  which is the last  p  in position 2  and then we are left with the center letter that needs no checking  When we do length 2 we stop at 1  so what happens is an extra check is performed with i being on the 1 position the  o   and compares it to itself  length-1-i       Checks if our string is palindromic  var ourString    A Man          amp  A Plan  A Canal  -Panama    isPalin ourString    function isPalin string       Make all lower case for case insensitivity and replace all spaces  underscores and non-words  string   string toLowerCase   replace   s  g      replace   W g     replace    g      for i 0  i lt  Math floor string length 2-1   i            if string i      string string length-1-i             console log  Your string is not palindromic             break          else if i     Math floor string length 2-1             console log  Your string is palindromic                       https   repl it FNVZ 1

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Assuming the leading zeros are ignored  Following is an implementation    include lt bits stdc   h gt  using namespace std  vector lt int gt digits  stack lt int gt digitsRev  int d number  bool isPal 1   initially assuming the number is palindrome int main         cin gt  gt number      if number lt 10   if it is a single digit number than it is palindrome               cout lt  lt  PALINDROME  lt  lt endl          return 0              if the number is greater than or equal to 10     while 1                d number 10   taking each digit         number number 10            vector and stack will pick the digits           in reverse order to each other         digits push back d           digitsRev push d           if number  0 break            int index 0      while  digitsRev empty                    Checking each element of the vector and the stack           to see if there is any inequality            And which is equivalent to check each digit of the main           number from both sides         if digitsRev top    digits index                           cout lt  lt  NOT PALINDROME  lt  lt endl              isPal 0              break                    digitsRev pop                If the digits are equal from both sides than the number is palindrome     if isPal  1 cout lt  lt  PALINDROME  lt  lt endl

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Just for fun  this one also works   a   num  b   0  if  a   10    0    return a    0  do     b   10   b   a   10    if  a    b      return true    a   a   10    while  a  gt  b   return a    b

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int is palindrome unsigned long orig        unsigned long reversed   0  n   orig       while  n  gt  0                reversed   reversed   10   n   10          n    10             return orig    reversed

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a method with a little better constant factor than  sminks method   num n lastDigit 0  rev 0  while  num gt rev        lastDigit num 10      rev rev 10 lastDigit      num   2    if  num  rev  print PALINDROME  exit 0   num num 10 lastDigit     This line is required as a number with odd number of bits will necessary end up being smaller even if it is a palindrome if  num  rev  print PALINDROME

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Push each individual digit onto a stack  then pop them off   If it s the same forwards and back  it s a palindrome

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Recursive solution in ruby  without converting the number to string   def palindrome  x  a x  b 0    return x  b if a lt 1   palindrome  x  a 10  b 10   a 10  end  palindrome  55655

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One line python code     def isPalindrome number       return True if str number        join reversed str number    else False

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This solution is quite efficient  since I am using StringBuilder which means that the StringBuilder Class is implemented as a mutable sequence of characters  This means that you append new Strings or chars onto a StringBuilder    public static boolean isPal String ss      StringBuilder stringBuilder   new StringBuilder ss      stringBuilder reverse       return ss equals stringBuilder toString

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A lot of the solutions posted here reverses the integer and stores it in a variable which uses extra space which is O n   but here is a solution with O 1  space   def isPalindrome num       if num  lt  0          return False     if num    0          return True     from math import log10     length   int log10 num       while length  gt  0          right   num   10         left   num   10  length         if right    left              return False         num    10  length         num    10         length -  2     return True

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Assuming the leading zeros are ignored  Following is an implementation    include lt bits stdc   h gt  using namespace std  vector lt int gt digits  stack lt int gt digitsRev  int d number  bool isPal 1   initially assuming the number is palindrome int main         cin gt  gt number      if number lt 10   if it is a single digit number than it is palindrome               cout lt  lt  PALINDROME  lt  lt endl          return 0              if the number is greater than or equal to 10     while 1                d number 10   taking each digit         number number 10            vector and stack will pick the digits           in reverse order to each other         digits push back d           digitsRev push d           if number  0 break            int index 0      while  digitsRev empty                    Checking each element of the vector and the stack           to see if there is any inequality            And which is equivalent to check each digit of the main           number from both sides         if digitsRev top    digits index                           cout lt  lt  NOT PALINDROME  lt  lt endl              isPal 0              break                    digitsRev pop                If the digits are equal from both sides than the number is palindrome     if isPal  1 cout lt  lt  PALINDROME  lt  lt endl

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Recursive solution in ruby  without converting the number to string   def palindrome  x  a x  b 0    return x  b if a lt 1   palindrome  x  a 10  b 10   a 10  end  palindrome  55655

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A lot of the solutions posted here reverses the integer and stores it in a variable which uses extra space which is O n   but here is a solution with O 1  space   def isPalindrome num       if num  lt  0          return False     if num    0          return True     from math import log10     length   int log10 num       while length  gt  0          right   num   10         left   num   10  length         if right    left              return False         num    10  length         num    10         length -  2     return True

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except making the number a string and then reversing the string    Why dismiss that solution  It s easy to implement and readable  If you were asked with no computer at hand whether 2  10-23 is a decimal palindrome  you d surely test it by writing it out in decimal   In Python  at least  the slogan  string operations are slower than arithmetic  is actually false  I compared Smink s arithmetical algorithm to simple string reversal int str i    -1    There was no significant difference in speed - it  happened string reversal was marginally faster   In compiled languages  C C    the slogan might hold  but one risks overflow errors with large numbers   def reverse n       rev   0     while n  gt  0          rev   rev   10   n   10         n   n    10     return rev  upper   10  6  def strung        for i in range upper           int str i    -1    def arithmetic        for i in range upper           reverse i   import timeit print  strung   timeit timeit  strung     setup  from   main   import strung   number 1  print  arithmetic   timeit timeit  arithmetic     setup  from   main   import arithmetic   number 1    Results in seconds  lower is better       strung 1 50960231881   arithmetic 1 69729960569

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I didn t notice any answers that solved this problem using no extra space  i e   all solutions I saw either used a string  or another integer to reverse the number  or some other data structures   Although languages like Java wrap around on integer overflow  this behavior is undefined in languages like C   Try reversing 2147483647  Integer MAX VALUE  in Java  Workaround could to be to use a long or something but  stylistically  I don t quite like that approach   Now  the concept of a palindromic number is that the number should read the same forwards and backwards  Great  Using this information  we can compare the first digit and the last digit  Trick is  for the first digit  we need the order of the number  Say  12321  Dividing this by 10000 would get us the leading 1  The trailing 1 can be retrieved by taking the mod with 10  Now  to reduce this to 232   12321   10000  10    2321  10   232  And now  the 10000 would need to be reduced by a factor of 2  So  now on to the Java code     private static boolean isPalindrome int n        if  n  lt  0          return false       int div   1         find the divisor     while  n   div  gt   10          div    10          any number less than 10 is a palindrome     while  n    0            int leading   n   div          int trailing   n   10          if  leading    trailing              return false                with div gets rid of leading digit            dividing result by 10 gets rid of trailing digit         n    n   div    10              got rid of 2 numbers  update div accordingly         div    100            return true      Edited as per Hardik s suggestion to cover the cases where there are zeroes in the number

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Try this       print     To Find Palindrome Number     def Palindrome Number                 n   input  Enter Number to check for palindromee                 m n              a   0        while m  0             a   m   10   a   10             m   m   10          if  n    a               print   d is a palindrome number   n      else          print   d is not a palindrome number   n    just call back the functions

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Golang version   package main  import  fmt   func main         n    123454321     r    reverse n      fmt Println r    n     func reverse n int  int       r    0     for           if n  gt  0               r   r 10   n 10             n   n   10           else               break                     return r

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def ReverseNumber n  partial 0       if n    0          return partial     return ReverseNumber n    10  partial   10   n   10   trial   123454321 if ReverseNumber trial     trial      print  It s a Palindrome      Works for integers only  It s unclear from the problem statement if floating point numbers or leading zeros need to be considered

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Here is a solution usings lists as stacks in python     def isPalindromicNum n                   is  n  a palindromic number              ns   list str n       for n in ns          if n    ns pop                return False     return True   popping the stack only considers the rightmost side of the number for comparison and it fails fast to reduce checks

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Try this   reverse   0      remainder   0      count   0      while  number  gt  reverse                remainder   number   10          reverse   reverse   10   remainder          number   number   10          count              Console WriteLine count       if  reverse    number                Console WriteLine  Your number is a palindrome              else               number   number   10   remainder          if  reverse    number              Console WriteLine  your number is a palindrome            else             Console WriteLine  your number is not a palindrome              Console ReadLine

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I didn t notice any answers that solved this problem using no extra space  i e   all solutions I saw either used a string  or another integer to reverse the number  or some other data structures   Although languages like Java wrap around on integer overflow  this behavior is undefined in languages like C   Try reversing 2147483647  Integer MAX VALUE  in Java  Workaround could to be to use a long or something but  stylistically  I don t quite like that approach   Now  the concept of a palindromic number is that the number should read the same forwards and backwards  Great  Using this information  we can compare the first digit and the last digit  Trick is  for the first digit  we need the order of the number  Say  12321  Dividing this by 10000 would get us the leading 1  The trailing 1 can be retrieved by taking the mod with 10  Now  to reduce this to 232   12321   10000  10    2321  10   232  And now  the 10000 would need to be reduced by a factor of 2  So  now on to the Java code     private static boolean isPalindrome int n        if  n  lt  0          return false       int div   1         find the divisor     while  n   div  gt   10          div    10          any number less than 10 is a palindrome     while  n    0            int leading   n   div          int trailing   n   10          if  leading    trailing              return false                with div gets rid of leading digit            dividing result by 10 gets rid of trailing digit         n    n   div    10              got rid of 2 numbers  update div accordingly         div    100            return true      Edited as per Hardik s suggestion to cover the cases where there are zeroes in the number

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Above most of the answers having a trivial problem is that the int variable possibly might overflow   Refer to http   articles leetcode com palindrome-number   boolean isPalindrome int x        if  x  lt  0          return false      int div   1      while  x   div  gt   10            div    10            while  x    0            int l   x   div          int r   x   10          if  l    r              return false          x    x   div    10          div    100            return true

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For any given number   n   num  rev   0  while  num  gt  0        dig   num   10      rev   rev   10   dig      num   num   10      If n    rev then num is a palindrome   cout  lt  lt   Number    lt  lt   n    rev    IS     IS NOT    lt  lt    a palindrome   lt  lt  endl

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Personally I do it this way  and there s no overlap  the code stops checking for matching characters in the correct spot whether the string has an even or odd length  Some of the other methods posted above will attempt to match one extra time when it doesn t need to    If we use length 2  it will still work but it will do one additional check that it doesn t need to do  For instance   pop  is 3 in length  3 2   1 5  so it will stop checking when i   2 since 1 lt 1 5 it will check when i   1 as well  but we need it to stop at 0  not one  The first  p  is in position 0  and it will check itself against length-1-0 current position  which is the last  p  in position 2  and then we are left with the center letter that needs no checking  When we do length 2 we stop at 1  so what happens is an extra check is performed with i being on the 1 position the  o   and compares it to itself  length-1-i       Checks if our string is palindromic  var ourString    A Man          amp  A Plan  A Canal  -Panama    isPalin ourString    function isPalin string       Make all lower case for case insensitivity and replace all spaces  underscores and non-words  string   string toLowerCase   replace   s  g      replace   W g     replace    g      for i 0  i lt  Math floor string length 2-1   i            if string i      string string length-1-i             console log  Your string is not palindromic             break          else if i     Math floor string length 2-1             console log  Your string is palindromic                       https   repl it FNVZ 1

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def palindrome n       d          while  n  gt  0           d append n   10          n     10     for i in range len d  2           if  d i     d - i 1                 return  Fail       return  Pass

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I answered the Euler problem using a very brute-forcy way  Naturally  there was a much smarter algorithm at display when I got to the new unlocked associated forum thread  Namely  a member who went by the handle Begoner had such a novel approach  that I decided to reimplement my solution using his algorithm  His version was in Python  using nested loops  and I reimplemented it in Clojure  using a single loop recur    Here for your amusement    defn palindrome   n     let  len  count n        and           first n   last n          or   gt   1  count n            palindrome     n  substring 1  dec len           defn begoners-palindrome       loop  mx 0          mxI 0          mxJ 0          i 999          j 990       if   gt  i 100         let  product    i j            if  and   gt  product mx   palindrome   str product               recur product i j              if   gt  j 100  i  dec i                if   gt  j 100   - j 11  990              recur mx mxI mxJ              if   gt  j 100  i  dec i                if   gt  j 100   - j 11  990           mx      time  prn  begoners-palindrome      There were Common Lisp answers as well  but they were ungrokable to me

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Recursive way  not very efficient  just provide an option   Python code   def isPalindrome num       size   len str num       demoninator   10   size-1      return isPalindromeHelper num  size  demoninator   def isPalindromeHelper num  size  demoninator          wrapper function  used in recursive        if size  lt  1          return True     else                 if num demoninator    num 10              return False           shrink the size  num and denominator         num    demoninator         num    10         size -  2         demoninator   100         return isPalindromeHelper num  size  demoninator

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I answered the Euler problem using a very brute-forcy way  Naturally  there was a much smarter algorithm at display when I got to the new unlocked associated forum thread  Namely  a member who went by the handle Begoner had such a novel approach  that I decided to reimplement my solution using his algorithm  His version was in Python  using nested loops  and I reimplemented it in Clojure  using a single loop recur    Here for your amusement    defn palindrome   n     let  len  count n        and           first n   last n          or   gt   1  count n            palindrome     n  substring 1  dec len           defn begoners-palindrome       loop  mx 0          mxI 0          mxJ 0          i 999          j 990       if   gt  i 100         let  product    i j            if  and   gt  product mx   palindrome   str product               recur product i j              if   gt  j 100  i  dec i                if   gt  j 100   - j 11  990              recur mx mxI mxJ              if   gt  j 100  i  dec i                if   gt  j 100   - j 11  990           mx      time  prn  begoners-palindrome      There were Common Lisp answers as well  but they were ungrokable to me

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One line python code     def isPalindrome number       return True if str number        join reversed str number    else False

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This solution is quite efficient  since I am using StringBuilder which means that the StringBuilder Class is implemented as a mutable sequence of characters  This means that you append new Strings or chars onto a StringBuilder    public static boolean isPal String ss      StringBuilder stringBuilder   new StringBuilder ss      stringBuilder reverse       return ss equals stringBuilder toString

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public boolean isPalindrome int x            if  isNegative x               return false           boolean isPalindrome   reverseNumber x     x   true   false          return isPalindrome             private boolean isNegative int x            if  x  lt  0              return true          return false             public int reverseNumber int x             int reverseNumber   0           while  x  gt  0                int remainder   x   10              reverseNumber   reverseNumber   10   remainder              x   x   10                     return reverseNumber

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def ReverseNumber n  partial 0       if n    0          return partial     return ReverseNumber n    10  partial   10   n   10   trial   123454321 if ReverseNumber trial     trial      print  It s a Palindrome      Works for integers only  It s unclear from the problem statement if floating point numbers or leading zeros need to be considered

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This is my Java code  Basically comparing the the first and last value of the string and next inner value pair and so forth         Palindrome number       String sNumber    12321       int l   sNumber length       getting the length of sNumber  In this case its 5     boolean flag   true      for  int i   0  i  lt   l    i            if  sNumber charAt i     sNumber charAt  l--  -1       comparing the first and the last values of the string             System out println sNumber    is not a palindrome number                flag   false              break                      l--     to reducing the length value by 1            if  flag            System out println sNumber    is a palindrome number

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A number is palindromic if its string representation is palindromic   def is palindrome s       return all s i     s - i   1   for i in range len s   2    def number palindrome n       return is palindrome str n

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This is one of the Project Euler problems   When I solved it in Haskell I did exactly what you suggest  convert the number to a String   It s then trivial to check that the string is a pallindrome   If it performs well enough  then why bother making it more complex   Being a pallindrome is a lexical property rather than a mathematical one

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public class Numbers       public static void main int givenNum               int n  givenNum        int rev 0          while n gt 0                       To extract the last digit           int digit n 10               To store it in reverse           rev  rev 10  digit               To throw the last digit           n n 10                   To check if a number is palindrome or not       if rev  givenNum                    System out println givenNum  is a palindrome                   else                  System out pritnln givenNum  is not a palindrome

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Dammit  I   m mad   http   www palindromelist net palindromes-d  Single steps example  is 332 a palindromic number   n   332 q   n   10   33 r   n - 10   q   2 r  gt  0 r    q n   q   33 n  gt  r q   n   10   3 r -  q   4294967295 r    10   4294967286 r    n   23 r    n r    q n   q   3 n  gt  r   No  so 332 isn t a palindromic number    Overflow isn t a problem either  Two divisions were necessary  in code  C   they re done with multiplications  A n-digit number   n 2 divisions   const ulong c0   0xcccccccdUL  static bool isPal uint n        if  n  lt  10  return true      uint q    uint  c0   n  gt  gt  35       uint r   n - 10   q      if  r    0  return false      if  r    q  return true      n   q      while  n  gt  r                q    uint  c0   n  gt  gt  35           r -  q          r    10          r    n          if  r    n    r    q  return true          n   q            return false      There are 142948 palindromic numbers  lt  2 32  their sum is 137275874705916   using System  class Program       static void Main       take a break               uint n  c  var sw   System Diagnostics Stopwatch StartNew            n    0u  c   0  sw Restart    while  n  gt  0  if  isPal0 n--   c            Console WriteLine sw Elapsed         c                       76 s         n    0u  c   0  sw Restart    while  n  gt  0  if  isPal1 n--   c            Console WriteLine sw Elapsed         c                       42 s         n    0u  c   0  sw Restart    while  n  gt  0  if  isPal2 n--   c            Console WriteLine sw Elapsed         c                       31 s         Console Read               static bool isPal0 uint u                uint n   u  rev   0          while  n  gt  0    uint dig   n   10  rev   rev   10   dig  n    10            return u    rev             static bool isPal1 uint u                uint n   u  r   0          while  n  gt   10  r   n    r -  n    10     10          return u    10   r   n             static bool isPal2 uint n                if  n  lt  10  return true          uint q   n   10  r   n - 10   q          if  r    0    r    q  return r  gt  0          while   n   q   gt  r                        q    10  r -  q  r    10  r    n              if  r    n    r    q  return true                    return false            This one seems to be faster   using System  class Program       static void Main                 uint n  c  var sw   System Diagnostics Stopwatch StartNew            n    0u  c   0  sw Restart    while  n  gt  0  if  isPal n--   c            Console WriteLine sw Elapsed         c                      21 s         Console Read               static bool isPal uint n                return n  lt  100   n  lt  10    n   11    0                 n  lt  1000                  n   100    n   10                n  lt  10000   n   11    0  amp  amp  n   1000    n   10  amp  amp  isP n                n  lt  100000                  n   10000    n   10  amp  amp  isP n               n  lt  1000000   n   11    0  amp  amp  n   100000    n   10  amp  amp  isP n              n  lt  10000000                  n   1000000    n   10  amp  amp  isP n             n  lt  100000000   n   11    0  amp  amp  n   10000000    n   10  amp  amp  isP n            n  lt  1000000000                  n   100000000    n   10  amp  amp  isP n                             n   11    0  amp  amp  n   1000000000    n   10  amp  amp  isP n              static bool isP uint n                uint q   n   10  r   n - 10   q          do   n   q  q    10  r -  q  r    10  r    n    while  r  lt  q           return r    q    r    n            With an almost balanced binary search spaghetti tree   using System  class Program       static void Main                 uint n  c  var sw   System Diagnostics Stopwatch StartNew            n   c   0  sw Restart    while  n  lt   0u  if  isPal n     c            Console WriteLine sw Elapsed         c                   17 s         Console Read               static bool isPal uint n                return n  lt  1000000   n  lt  10000   n  lt  1000   n  lt  100           n  lt  10    n   11    0   n   100    n   10           n   1000    n   10  amp  amp  isP n    n  lt  100000           n   10000    n   10  amp  amp  isP n            n   100000    n   10  amp  amp  isP n            n  lt  100000000   n  lt  10000000           n   1000000    n   10  amp  amp  isP n            n   11    0  amp  amp  n   10000000    n   10  amp  amp  isP n            n  lt  1000000000   n   100000000    n   10  amp  amp  isP n            n   11    0  amp  amp  n   1000000000    n   10  amp  amp  isP n              static bool isP uint n                uint q   n   10  r   n - 10   q          do   n   q  q    10  r -  q  r    10  r    n    while  r  lt  q           return r    q    r    n            Unbalanced upside down   using System  class Program       static void Main                 uint n  c  var sw   System Diagnostics Stopwatch StartNew            n   c   0  sw Restart    while  n  lt   0u  if  isPal n     c            Console WriteLine sw Elapsed         c                   16 s         Console Read               static bool isPal uint n                return         n  gt  999999999   n   11    0  amp  amp  n   1000000000    n   10  amp  amp  isP n            n  gt  99999999   n   100000000    n   10  amp  amp  isP n            n  gt  9999999   n   11    0  amp  amp  n   10000000    n   10  amp  amp  isP n            n  gt  999999   n   1000000    n   10  amp  amp  isP n            n  gt  99999   n   11    0  amp  amp  n   100000    n   10  amp  amp  isP n            n  gt  9999   n   10000    n   10  amp  amp  isP n            n  gt  999   n   11    0  amp  amp  n   1000    n   10  amp  amp  isP n            n  gt  99   n   100    n   10           n  lt  10    n   11    0             static bool isP uint n                uint q   n   10  r   n - 10   q          do   n   q  q    10  r -  q  r    10  r    n    while  r  lt  q           return r    q    r    n

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checkPalindrome int number        int lsd  msd len      len   log10 number       while number                msd    number pow 10 len        most significant digit          lsd   number 10      least significant digit          if lsd  msd                        number  10     change of LSD             number- msd pow 10 --len      change of MSD  due to change of MSD             len- 1     due to change in LSD               else  return 1             return 0

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Recursive way  not very efficient  just provide an option   Python code   def isPalindrome num       size   len str num       demoninator   10   size-1      return isPalindromeHelper num  size  demoninator   def isPalindromeHelper num  size  demoninator          wrapper function  used in recursive        if size  lt  1          return True     else                 if num demoninator    num 10              return False           shrink the size  num and denominator         num    demoninator         num    10         size -  2         demoninator   100         return isPalindromeHelper num  size  demoninator

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In Python  there is a fast  iterative way   def reverse n       newnum 0     while n gt 0          newnum   newnum 10   n   10         n   10     return newnum  def palindrome n       return n    reverse n    This also prevents memory issues with recursion  like StackOverflow error in Java

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This is one of the Project Euler problems   When I solved it in Haskell I did exactly what you suggest  convert the number to a String   It s then trivial to check that the string is a pallindrome   If it performs well enough  then why bother making it more complex   Being a pallindrome is a lexical property rather than a mathematical one

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public class PalindromePrime       private static int g  n  0 i m      private javax swing JTextField jTextField      static String b      private static Scanner scanner   new Scanner  System in      public static void main String    args  throws IOException         System out print   Please Inter Data            g   scanner nextInt          System out print   Please Inter Data 2             m   scanner nextInt           count g m                  public static    int count int L  int R        int resultNum   0       for  i  L   i lt   R  i             int count  0           for  n   i   n  gt  1  n --                 if i n  0                                count   count   1                   System out println    Data    n     count                                               if count    2                             b   b  i                 String ss   String  valueOf i               System out print   n   i                      if isPalindrome i                                0 System out println  Number       i     is   a palindrome                           number2 b    Integer parseInt number ayy b                       String s   String  valueOf i                     System out printf  123456   s                   resultNum                               else                      System out println  Number       i     is Not  a palindrome                                System out println    Data    n    ss length                              palindrome i               System out print    Data             System out println    Data    n     b         return resultNum             SuppressWarnings  unused   public static boolean isPalindrome int number          int p   number             p      int            number                  method      int r   0           r      int                             0      int w   0       while  p    0                 While     p            0        2   0                       w   p   10                   W                p          parramiter      amp  mod      10           2   10   2   w  2   3  10   w  3        r   r   10   w            R                                   10    w             w   p   10      mod           0 10   2   2         p   p   10             p              paramiter            10                                                                0                                       1                 p    0                 p       p   number                  2 r    r   10     p 10             3   p   p  10                            0           Loop      if  number    r               for int count   0   count  lt i  count          String s1   String valueOf i                   countLines s1           System out println  Number           s1         is   a palindrome             return true          return                return false     public static int countLines String str        if  str    null    str length      0          return 0      int lines   1      int len   str length        for  int pos   0  pos  lt  len  pos              char c   str charAt pos           if  c      r                    System out println  Line 0           str                 lines                if   pos 1  lt  len  amp  amp  str charAt pos 1       n                     System out println  Line           str                     pos              else if  c      n                  lines                 System out println  Line 2           str                         return lines      public static int countLines1 String sd  throws IOException       LineNumberReader lineNumberReader   new LineNumberReader new StringReader sd        int count   0       System out printf  Line        count   count   1        lineNumberReader skip Long MAX VALUE       return lineNumberReader getLineNumber

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let isPalindrome  n int       let l1   n ToString     gt  List ofSeq   gt  List rev    let rec isPalindromeInt l1 l2          match  l1 l2  with           h1  rest1 h2  rest2  - gt  if  h1   h2  then isPalindromeInt rest1 rest2 else false            - gt  true    isPalindromeInt l1  n ToString     gt  List ofSeq

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Just for fun  this one also works   a   num  b   0  if  a   10    0    return a    0  do     b   10   b   a   10    if  a    b      return true    a   a   10    while  a  gt  b   return a    b

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To check the given number is Palindrome or not  Java Code   class CheckPalindrome  public static void main String str             int a 242  n a  b a  rev 0          while n gt 0                       a n 10   n n 10 rev rev 10 a                      System out println a      n      rev       to see the logic                          if rev  b   System out println  Palindrome            else        System out println  Not Palindrome

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For any given number   n   num  rev   0  while  num  gt  0        dig   num   10      rev   rev   10   dig      num   num   10      If n    rev then num is a palindrome   cout  lt  lt   Number    lt  lt   n    rev    IS     IS NOT    lt  lt    a palindrome   lt  lt  endl

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Check this solution in java     private static boolean ispalidrome long n            return getrev n  0L     n             private static long getrev long n  long initvalue            if  n  lt   0                return initvalue                    initvalue    initvalue   10     n   10           return getrev n   10  initvalue

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public static boolean isPalindrome int x            int newX   x          int newNum   0          boolean result   false          if  x  gt   0                while  newX  gt   10                    newNum   newNum newX   10                  newNum   newNum   10                  newX   newX   10                            newNum    newX               if newNum  x                result   true                            else                   result false                                   else                result   false                    return result

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In Python  there is a fast  iterative way   def reverse n       newnum 0     while n gt 0          newnum   newnum 10   n   10         n   10     return newnum  def palindrome n       return n    reverse n    This also prevents memory issues with recursion  like StackOverflow error in Java

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I always use this python solution due to its compactness   def isPalindrome number       return int str number    -1    number

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Try this   reverse   0      remainder   0      count   0      while  number  gt  reverse                remainder   number   10          reverse   reverse   10   remainder          number   number   10          count              Console WriteLine count       if  reverse    number                Console WriteLine  Your number is a palindrome              else               number   number   10   remainder          if  reverse    number              Console WriteLine  your number is a palindrome            else             Console WriteLine  your number is not a palindrome              Console ReadLine

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num   int raw input    list num   list str num   if list num   -1     list num      print  Its a palindrome  else      print  Its not a palindrom

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Fastest way I know   bool is pal int n        if  n   10    0  return 0      int r   0      while  r  lt  n            r   10   r   n   10          n    10            return n    r    n    r   10

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Here is a way   class Palindrome Number      void display int a           int count 0          int n a          int n1 a          while a gt 0               count                a a 10                    double b 0 0d          while n gt 0               b   n 10   Math pow 10 count-1                count--              n n 10                    if b   double n1               System out println  Palindrome number                      else              System out println  Not a palindrome number

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except making the number a string and then reversing the string    Why dismiss that solution  It s easy to implement and readable  If you were asked with no computer at hand whether 2  10-23 is a decimal palindrome  you d surely test it by writing it out in decimal   In Python  at least  the slogan  string operations are slower than arithmetic  is actually false  I compared Smink s arithmetical algorithm to simple string reversal int str i    -1    There was no significant difference in speed - it  happened string reversal was marginally faster   In compiled languages  C C    the slogan might hold  but one risks overflow errors with large numbers   def reverse n       rev   0     while n  gt  0          rev   rev   10   n   10         n   n    10     return rev  upper   10  6  def strung        for i in range upper           int str i    -1    def arithmetic        for i in range upper           reverse i   import timeit print  strung   timeit timeit  strung     setup  from   main   import strung   number 1  print  arithmetic   timeit timeit  arithmetic     setup  from   main   import arithmetic   number 1    Results in seconds  lower is better       strung 1 50960231881   arithmetic 1 69729960569

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Here is an Scheme version that constructs a function that will work against any base  It has a redundancy check  return false quickly if the number is a multiple of the base  ends in 0   And it doesn t rebuild the entire reversed number  only half  That s all we need    define make-palindrome-tester     lambda  base        lambda  n          cond              0  modulo n base    f            else            letrec                 Q  lambda  h t                       cond                          lt  h t   f                            h t   t                         else                         let                               h2  quotient h base                                m   - h    h2 base                               cond                                h2 t   t                              else                              Q h2       base t  m                        Q n 0

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public static void main String args          System out print  Enter a number          Scanner input   new Scanner System in       int num   input nextInt        int number   num      int reversenum   0      while  num    0                reversenum   reversenum   10          reversenum   reversenum   num   10          num   num   10             if  number    reversenum          System out println  The reverse number is     reversenum     nThen the number is palindrome         else         System out println  The reverse number is     reversenum     nThen the number is not palindrome

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public static boolean isPalindrome int x            int newX   x          int newNum   0          boolean result   false          if  x  gt   0                while  newX  gt   10                    newNum   newNum newX   10                  newNum   newNum   10                  newX   newX   10                            newNum    newX               if newNum  x                result   true                            else                   result false                                   else                result   false                    return result

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Pop off the first and last digits and compare them until you run out   There may be a digit left  or not  but either way  if all the popped off digits match  it is a palindrome

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public class PalindromePrime       private static int g  n  0 i m      private javax swing JTextField jTextField      static String b      private static Scanner scanner   new Scanner  System in      public static void main String    args  throws IOException         System out print   Please Inter Data            g   scanner nextInt          System out print   Please Inter Data 2             m   scanner nextInt           count g m                  public static    int count int L  int R        int resultNum   0       for  i  L   i lt   R  i             int count  0           for  n   i   n  gt  1  n --                 if i n  0                                count   count   1                   System out println    Data    n     count                                               if count    2                             b   b  i                 String ss   String  valueOf i               System out print   n   i                      if isPalindrome i                                0 System out println  Number       i     is   a palindrome                           number2 b    Integer parseInt number ayy b                       String s   String  valueOf i                     System out printf  123456   s                   resultNum                               else                      System out println  Number       i     is Not  a palindrome                                System out println    Data    n    ss length                              palindrome i               System out print    Data             System out println    Data    n     b         return resultNum             SuppressWarnings  unused   public static boolean isPalindrome int number          int p   number             p      int            number                  method      int r   0           r      int                             0      int w   0       while  p    0                 While     p            0        2   0                       w   p   10                   W                p          parramiter      amp  mod      10           2   10   2   w  2   3  10   w  3        r   r   10   w            R                                   10    w             w   p   10      mod           0 10   2   2         p   p   10             p              paramiter            10                                                                0                                       1                 p    0                 p       p   number                  2 r    r   10     p 10             3   p   p  10                            0           Loop      if  number    r               for int count   0   count  lt i  count          String s1   String valueOf i                   countLines s1           System out println  Number           s1         is   a palindrome             return true          return                return false     public static int countLines String str        if  str    null    str length      0          return 0      int lines   1      int len   str length        for  int pos   0  pos  lt  len  pos              char c   str charAt pos           if  c      r                    System out println  Line 0           str                 lines                if   pos 1  lt  len  amp  amp  str charAt pos 1       n                     System out println  Line           str                     pos              else if  c      n                  lines                 System out println  Line 2           str                         return lines      public static int countLines1 String sd  throws IOException       LineNumberReader lineNumberReader   new LineNumberReader new StringReader sd        int count   0       System out printf  Line        count   count   1        lineNumberReader skip Long MAX VALUE       return lineNumberReader getLineNumber

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Just for fun  this one also works   a   num  b   0  if  a   10    0    return a    0  do     b   10   b   a   10    if  a    b      return true    a   a   10    while  a  gt  b   return a    b

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a method with a little better constant factor than  sminks method   num n lastDigit 0  rev 0  while  num gt rev        lastDigit num 10      rev rev 10 lastDigit      num   2    if  num  rev  print PALINDROME  exit 0   num num 10 lastDigit     This line is required as a number with odd number of bits will necessary end up being smaller even if it is a palindrome if  num  rev  print PALINDROME

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Pop off the first and last digits and compare them until you run out   There may be a digit left  or not  but either way  if all the popped off digits match  it is a palindrome

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Try this       print     To Find Palindrome Number     def Palindrome Number                 n   input  Enter Number to check for palindromee                 m n              a   0        while m  0             a   m   10   a   10             m   m   10          if  n    a               print   d is a palindrome number   n      else          print   d is not a palindrome number   n    just call back the functions

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This code converts int to String and then checks if the string is pallindrome  The advantage is that it is fast  the disadvantage being that it converts int to String thereby compromising with the perfect solution to question    static int pallindrome 41012  static String pallindromer  Integer toString pallindrome    static int length pallindromer length     public static void main String   args        pallindrome 0       System out println  It s a pallindrome       static void pallindrome int index       if pallindromer charAt index   pallindromer charAt length- index 1             if index lt length-1               pallindrome   index                       else          System out println  Not a pallindrome            System exit 0

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This is one of the Project Euler problems   When I solved it in Haskell I did exactly what you suggest  convert the number to a String   It s then trivial to check that the string is a pallindrome   If it performs well enough  then why bother making it more complex   Being a pallindrome is a lexical property rather than a mathematical one

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Above most of the answers having a trivial problem is that the int variable possibly might overflow   Refer to http   articles leetcode com palindrome-number   boolean isPalindrome int x        if  x  lt  0          return false      int div   1      while  x   div  gt   10            div    10            while  x    0            int l   x   div          int r   x   10          if  l    r              return false          x    x   div    10          div    100            return true

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Below is the answer in swift   It reads number from left and right side and compare them if they are same  Doing this way we will never face a problem of integer overflow  which can occure on reversing number method  as we are not creating another number    Steps    Get length of number Loop from length   1 first  --  0 Get ith digit  amp  get last digit if both digits are not equal return false as number is not palindrome i -- discard last digit from num  num   num   10  end of loo return true  func isPalindrom   input  Int  - gt  Bool          if input  lt  0               return false                    if input  lt  10               return true                    var num   input         let length   Int log10 Float input      1         var i   length          while i  gt  0  amp  amp  num  gt  0                let ithDigit    input   Int pow 10 0  Double i  - 1 0       10             let r   Int num   10               if ithDigit    r                   return false                            num   num   10             i -  1                    return true

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Golang version   package main  import  fmt   func main         n    123454321     r    reverse n      fmt Println r    n     func reverse n int  int       r    0     for           if n  gt  0               r   r 10   n 10             n   n   10           else               break                     return r

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Here is a solution usings lists as stacks in python     def isPalindromicNum n                   is  n  a palindromic number              ns   list str n       for n in ns          if n    ns pop                return False     return True   popping the stack only considers the rightmost side of the number for comparison and it fails fast to reduce checks

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I answered the Euler problem using a very brute-forcy way  Naturally  there was a much smarter algorithm at display when I got to the new unlocked associated forum thread  Namely  a member who went by the handle Begoner had such a novel approach  that I decided to reimplement my solution using his algorithm  His version was in Python  using nested loops  and I reimplemented it in Clojure  using a single loop recur    Here for your amusement    defn palindrome   n     let  len  count n        and           first n   last n          or   gt   1  count n            palindrome     n  substring 1  dec len           defn begoners-palindrome       loop  mx 0          mxI 0          mxJ 0          i 999          j 990       if   gt  i 100         let  product    i j            if  and   gt  product mx   palindrome   str product               recur product i j              if   gt  j 100  i  dec i                if   gt  j 100   - j 11  990              recur mx mxI mxJ              if   gt  j 100  i  dec i                if   gt  j 100   - j 11  990           mx      time  prn  begoners-palindrome      There were Common Lisp answers as well  but they were ungrokable to me

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Dammit  I   m mad   http   www palindromelist net palindromes-d  Single steps example  is 332 a palindromic number   n   332 q   n   10   33 r   n - 10   q   2 r  gt  0 r    q n   q   33 n  gt  r q   n   10   3 r -  q   4294967295 r    10   4294967286 r    n   23 r    n r    q n   q   3 n  gt  r   No  so 332 isn t a palindromic number    Overflow isn t a problem either  Two divisions were necessary  in code  C   they re done with multiplications  A n-digit number   n 2 divisions   const ulong c0   0xcccccccdUL  static bool isPal uint n        if  n  lt  10  return true      uint q    uint  c0   n  gt  gt  35       uint r   n - 10   q      if  r    0  return false      if  r    q  return true      n   q      while  n  gt  r                q    uint  c0   n  gt  gt  35           r -  q          r    10          r    n          if  r    n    r    q  return true          n   q            return false      There are 142948 palindromic numbers  lt  2 32  their sum is 137275874705916   using System  class Program       static void Main       take a break               uint n  c  var sw   System Diagnostics Stopwatch StartNew            n    0u  c   0  sw Restart    while  n  gt  0  if  isPal0 n--   c            Console WriteLine sw Elapsed         c                       76 s         n    0u  c   0  sw Restart    while  n  gt  0  if  isPal1 n--   c            Console WriteLine sw Elapsed         c                       42 s         n    0u  c   0  sw Restart    while  n  gt  0  if  isPal2 n--   c            Console WriteLine sw Elapsed         c                       31 s         Console Read               static bool isPal0 uint u                uint n   u  rev   0          while  n  gt  0    uint dig   n   10  rev   rev   10   dig  n    10            return u    rev             static bool isPal1 uint u                uint n   u  r   0          while  n  gt   10  r   n    r -  n    10     10          return u    10   r   n             static bool isPal2 uint n                if  n  lt  10  return true          uint q   n   10  r   n - 10   q          if  r    0    r    q  return r  gt  0          while   n   q   gt  r                        q    10  r -  q  r    10  r    n              if  r    n    r    q  return true                    return false            This one seems to be faster   using System  class Program       static void Main                 uint n  c  var sw   System Diagnostics Stopwatch StartNew            n    0u  c   0  sw Restart    while  n  gt  0  if  isPal n--   c            Console WriteLine sw Elapsed         c                      21 s         Console Read               static bool isPal uint n                return n  lt  100   n  lt  10    n   11    0                 n  lt  1000                  n   100    n   10                n  lt  10000   n   11    0  amp  amp  n   1000    n   10  amp  amp  isP n                n  lt  100000                  n   10000    n   10  amp  amp  isP n               n  lt  1000000   n   11    0  amp  amp  n   100000    n   10  amp  amp  isP n              n  lt  10000000                  n   1000000    n   10  amp  amp  isP n             n  lt  100000000   n   11    0  amp  amp  n   10000000    n   10  amp  amp  isP n            n  lt  1000000000                  n   100000000    n   10  amp  amp  isP n                             n   11    0  amp  amp  n   1000000000    n   10  amp  amp  isP n              static bool isP uint n                uint q   n   10  r   n - 10   q          do   n   q  q    10  r -  q  r    10  r    n    while  r  lt  q           return r    q    r    n            With an almost balanced binary search spaghetti tree   using System  class Program       static void Main                 uint n  c  var sw   System Diagnostics Stopwatch StartNew            n   c   0  sw Restart    while  n  lt   0u  if  isPal n     c            Console WriteLine sw Elapsed         c                   17 s         Console Read               static bool isPal uint n                return n  lt  1000000   n  lt  10000   n  lt  1000   n  lt  100           n  lt  10    n   11    0   n   100    n   10           n   1000    n   10  amp  amp  isP n    n  lt  100000           n   10000    n   10  amp  amp  isP n            n   100000    n   10  amp  amp  isP n            n  lt  100000000   n  lt  10000000           n   1000000    n   10  amp  amp  isP n            n   11    0  amp  amp  n   10000000    n   10  amp  amp  isP n            n  lt  1000000000   n   100000000    n   10  amp  amp  isP n            n   11    0  amp  amp  n   1000000000    n   10  amp  amp  isP n              static bool isP uint n                uint q   n   10  r   n - 10   q          do   n   q  q    10  r -  q  r    10  r    n    while  r  lt  q           return r    q    r    n            Unbalanced upside down   using System  class Program       static void Main                 uint n  c  var sw   System Diagnostics Stopwatch StartNew            n   c   0  sw Restart    while  n  lt   0u  if  isPal n     c            Console WriteLine sw Elapsed         c                   16 s         Console Read               static bool isPal uint n                return         n  gt  999999999   n   11    0  amp  amp  n   1000000000    n   10  amp  amp  isP n            n  gt  99999999   n   100000000    n   10  amp  amp  isP n            n  gt  9999999   n   11    0  amp  amp  n   10000000    n   10  amp  amp  isP n            n  gt  999999   n   1000000    n   10  amp  amp  isP n            n  gt  99999   n   11    0  amp  amp  n   100000    n   10  amp  amp  isP n            n  gt  9999   n   10000    n   10  amp  amp  isP n            n  gt  999   n   11    0  amp  amp  n   1000    n   10  amp  amp  isP n            n  gt  99   n   100    n   10           n  lt  10    n   11    0             static bool isP uint n                uint q   n   10  r   n - 10   q          do   n   q  q    10  r -  q  r    10  r    n    while  r  lt  q           return r    q    r    n

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here s a f  version   let reverseNumber n       let rec loop acc   function      0 - gt  acc      x - gt  loop  acc   10   x   10   x 10          loop 0 n  let isPalindrome   function       x  when x   reverseNumber x - gt  true         - gt  false

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I went with the regular approach by converting number to string and then further converting string to charArray  Traversing through charArray and find if number at positions are equal or not  Note-  Not reversing the string   public bool IsPalindrome int num                string st   num ToString            char   arr   st ToCharArray            int len   arr Length          if  len  lt   1                        return false                    for  int i   0  i  lt  arr Length  i                          if  arr i     arr len - 1                                 if  i  gt   len                                        return true                                    len--                            else                               break                                  return false

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Simply get the digit count of the number via Math functions and then iterate by using     and     operations as follows  After x    x   divider    10  we should divide divider by 100 since we made 2 zero operations   public static boolean isPalindrome int x                if  x  lt  0  return false              if  x  lt  10  return true               int numDigits    int  Math log10 x  1               int divider    int   Math pow 10  numDigits - 1                for  int i   0  i  lt  numDigits   2  i                      if  x   divider    x   10                      return false                  x    x   divider    10                  divider    100                            return true

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Push each individual digit onto a stack  then pop them off   If it s the same forwards and back  it s a palindrome

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To check the given number is Palindrome or not  Java Code   class CheckPalindrome  public static void main String str             int a 242  n a  b a  rev 0          while n gt 0                       a n 10   n n 10 rev rev 10 a                      System out println a      n      rev       to see the logic                          if rev  b   System out println  Palindrome            else        System out println  Not Palindrome

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int is palindrome unsigned long orig        unsigned long reversed   0  n   orig       while  n  gt  0                reversed   reversed   10   n   10          n    10             return orig    reversed

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it seems like the easest thing would be to find the opposit number and compare the two   int max   int  Math random   100001        int i      int num   max    a var used in the tests     int size    the number of digits in the original number     int opos   0     the oposite number     int nsize   1       System out println max        for i   1  num gt 10  i                  num   num 10             System out println  this number has   i   digits         size   i    setting the digit number to a var for later use        num   max       for i 1 i lt size i                  nsize   10              while num gt 1                opos     num 10  nsize          num  10          nsize  10             System out println  and the number backwards is   opos        if  opos    max                 System out println  palindrome                else               System out println  aint no palindrome

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checkPalindrome int number        int lsd  msd len      len   log10 number       while number                msd    number pow 10 len        most significant digit          lsd   number 10      least significant digit          if lsd  msd                        number  10     change of LSD             number- msd pow 10 --len      change of MSD  due to change of MSD             len- 1     due to change in LSD               else  return 1             return 0

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int is palindrome unsigned long orig        unsigned long reversed   0  n   orig       while  n  gt  0                reversed   reversed   10   n   10          n    10             return orig    reversed

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def ReverseNumber n  partial 0       if n    0          return partial     return ReverseNumber n    10  partial   10   n   10   trial   123454321 if ReverseNumber trial     trial      print  It s a Palindrome      Works for integers only  It s unclear from the problem statement if floating point numbers or leading zeros need to be considered

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I answered the Euler problem using a very brute-forcy way  Naturally  there was a much smarter algorithm at display when I got to the new unlocked associated forum thread  Namely  a member who went by the handle Begoner had such a novel approach  that I decided to reimplement my solution using his algorithm  His version was in Python  using nested loops  and I reimplemented it in Clojure  using a single loop recur    Here for your amusement    defn palindrome   n     let  len  count n        and           first n   last n          or   gt   1  count n            palindrome     n  substring 1  dec len           defn begoners-palindrome       loop  mx 0          mxI 0          mxJ 0          i 999          j 990       if   gt  i 100         let  product    i j            if  and   gt  product mx   palindrome   str product               recur product i j              if   gt  j 100  i  dec i                if   gt  j 100   - j 11  990              recur mx mxI mxJ              if   gt  j 100  i  dec i                if   gt  j 100   - j 11  990           mx      time  prn  begoners-palindrome      There were Common Lisp answers as well  but they were ungrokable to me

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it seems like the easest thing would be to find the opposit number and compare the two   int max   int  Math random   100001        int i      int num   max    a var used in the tests     int size    the number of digits in the original number     int opos   0     the oposite number     int nsize   1       System out println max        for i   1  num gt 10  i                  num   num 10             System out println  this number has   i   digits         size   i    setting the digit number to a var for later use        num   max       for i 1 i lt size i                  nsize   10              while num gt 1                opos     num 10  nsize          num  10          nsize  10             System out println  and the number backwards is   opos        if  opos    max                 System out println  palindrome                else               System out println  aint no palindrome

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Here is one more solution in c   using templates   This solution will work for case insensitive palindrome string comparison    template  lt typename bidirection iter gt  bool palindrome bidirection iter first  bidirection iter last        while first    last  amp  amp  first    --last                if   toupper  first       toupper  last               return false          else             first              return true

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Pop off the first and last digits and compare them until you run out   There may be a digit left  or not  but either way  if all the popped off digits match  it is a palindrome

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I went with the regular approach by converting number to string and then further converting string to charArray  Traversing through charArray and find if number at positions are equal or not  Note-  Not reversing the string   public bool IsPalindrome int num                string st   num ToString            char   arr   st ToCharArray            int len   arr Length          if  len  lt   1                        return false                    for  int i   0  i  lt  arr Length  i                          if  arr i     arr len - 1                                 if  i  gt   len                                        return true                                    len--                            else                               break                                  return false

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Here is one more solution in c   using templates   This solution will work for case insensitive palindrome string comparison    template  lt typename bidirection iter gt  bool palindrome bidirection iter first  bidirection iter last        while first    last  amp  amp  first    --last                if   toupper  first       toupper  last               return false          else             first              return true

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Fastest way I know   bool is pal int n        if  n   10    0  return 0      int r   0      while  r  lt  n            r   10   r   n   10          n    10            return n    r    n    r   10

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Not sure if this is the best answer  and I m new at programming   it s in java   import java util     public class Palin        public static void main String   args            Random randInt   new Random             Scanner kbd   new Scanner System in           int t   kbd nextInt        of test cases          String   arrPalin   new String t     array of inputs          String   nextPalin   new String t           for  int i   0  i  lt  t  i                  arrPalin i    String valueOf randInt nextInt 2147483646    1               System out println arrPalin i                       final long startTime   System nanoTime             for  int i   0  i  lt  t  i                  nextPalin i     unmatcher incrementer switcher match arrPalin i                           final long duration   System nanoTime   - startTime           for  int i   0  i  lt  t  i                  System out println nextPalin i                       System out println duration               public static String match String N            int length   N length               Initialize a string with length of N         char   chars   new char length           Arrays fill chars   0             int count   1           for  int i   0  i  lt  length  i                  if   i 2     0      at i   even                  if  i    0                        chars i    N charAt i                     else                     chars i    N charAt i 2                 else   at i   odd                 chars i    N charAt length - count                   count                       return String valueOf chars              public static String switcher String N            int length   N length            char   chars   new char length           Arrays fill chars   0             for  int i   0  i  lt  length  i                  if  i    0                    if   i   2     0                        chars i    N charAt i                     else if   i   2     0                        chars i    N charAt i - 1                                               if  i    0                    chars 0    N charAt 0                                   return String valueOf chars              public static String incrementer String N            int length   N length            char   chars   new char length           Arrays fill chars   0             char   newN   N toCharArray             String returnVal           int numOne  numTwo           if   length   2     0                numOne   N charAt length-1               numTwo   N charAt length-2               newN length-1     char  numOne 1               newN length-2     char  numTwo 1               returnVal   String valueOf newN             else               numOne   N charAt length-1               newN length-1     char  numOne 1               returnVal   String valueOf newN                     return returnVal             public static String unmatcher String N            int length   N length            char   chars   new char length           Arrays fill chars   0            char   newN   N toCharArray             for  int i   0  i  lt  length  i                      if    i   2     0   amp  amp   i    0        for i  gt  0  even                     newN i   2    N charAt i                     else if   i   2     0  amp  amp   i    0        for i   0                     newN 0    N charAt 0                                           for  int i    length 2   i  lt  length  i                  newN i    newN Math abs i -  length - 1                        return String valueOf newN               So for this code  input a number  I did random numbers of how many I want    it will convert  for example the input is 172345 to 157423  Then it will change it to 117722  then if even make it 117733  if odd do the same for the only the last digit  Then make it 173371  It s not specifically finding a palindrome  but it s finding the next highest palindrome number

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A number is palindromic if its string representation is palindromic   def is palindrome s       return all s i     s - i   1   for i in range len s   2    def number palindrome n       return is palindrome str n

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Simply get the digit count of the number via Math functions and then iterate by using     and     operations as follows  After x    x   divider    10  we should divide divider by 100 since we made 2 zero operations   public static boolean isPalindrome int x                if  x  lt  0  return false              if  x  lt  10  return true               int numDigits    int  Math log10 x  1               int divider    int   Math pow 10  numDigits - 1                for  int i   0  i  lt  numDigits   2  i                      if  x   divider    x   10                      return false                  x    x   divider    10                  divider    100                            return true

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Pop off the first and last digits and compare them until you run out   There may be a digit left  or not  but either way  if all the popped off digits match  it is a palindrome

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int reverse int num        assert num  gt   0        for non-negative integers only      int rev   0      while  num    0                rev   rev   10   num   10          num    10            return rev      This seemed to work too  but did you consider the possibility that the reversed number might overflow  If it overflows  the behavior is language specific  For Java the number wraps around on overflow  but in C C   its behavior is undefined   Yuck   It turns out that comparing from the two ends is easier  First  compare the first and last digit  If they are not the same  it must not be a palindrome  If they are the same  chop off one digit from both ends and continue until you have no digits left  which you conclude that it must be a palindrome   Now  getting and chopping the last digit is easy  However  getting and chopping the first digit in a generic way requires some thought  The solution below takes care of it   int isIntPalindrome int x        if  x  lt  0      return 0      int div   1      while  x   div  gt   10                div    10             while  x    0                int l   x   div          int r   x   10          if  l    r              return 0          x    x   div    10          div    100            return 1

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Here is a way   class Palindrome Number      void display int a           int count 0          int n a          int n1 a          while a gt 0               count                a a 10                    double b 0 0d          while n gt 0               b   n 10   Math pow 10 count-1                count--              n n 10                    if b   double n1               System out println  Palindrome number                      else              System out println  Not a palindrome number

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Below is the answer in swift   It reads number from left and right side and compare them if they are same  Doing this way we will never face a problem of integer overflow  which can occure on reversing number method  as we are not creating another number    Steps    Get length of number Loop from length   1 first  --  0 Get ith digit  amp  get last digit if both digits are not equal return false as number is not palindrome i -- discard last digit from num  num   num   10  end of loo return true  func isPalindrom   input  Int  - gt  Bool          if input  lt  0               return false                    if input  lt  10               return true                    var num   input         let length   Int log10 Float input      1         var i   length          while i  gt  0  amp  amp  num  gt  0                let ithDigit    input   Int pow 10 0  Double i  - 1 0       10             let r   Int num   10               if ithDigit    r                   return false                            num   num   10             i -  1                    return true

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This code converts int to String and then checks if the string is pallindrome  The advantage is that it is fast  the disadvantage being that it converts int to String thereby compromising with the perfect solution to question    static int pallindrome 41012  static String pallindromer  Integer toString pallindrome    static int length pallindromer length     public static void main String   args        pallindrome 0       System out println  It s a pallindrome       static void pallindrome int index       if pallindromer charAt index   pallindromer charAt length- index 1             if index lt length-1               pallindrome   index                       else          System out println  Not a pallindrome            System exit 0

[algorithm] How do I check if a number is a palindrome?

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