Java reverse an int value without using array

Question

Can anyone explain to me how to reverse an integer without using array or String  I got this code from online  but not really understand why   input   10 and divide again   while  input    0        reversedNum   reversedNum   10   input   10      input   input   10         And how to do use this sample code to reverse only odd number  Example I got this input 12345  then it will reverse the odd number to output 531

User · Answer

If the idea is not to use arrays or string, reversing an integer has to be done by reading the digits of a number from the end one at a time. Below explanation is provided in detail to help the novice.

pseudocode :

lets start with reversed_number = 0 and some value for original_number which needs to be reversed.
the_last_digit = original_number % 10 (i.e, the reminder after dividing by 10)
original_number = original_number/10 (since we already have the last digit, remove the last digit from the original_number)
reversed_number = reversed_number * 10 + last_digit (multiply the reversed_number with 10, so as to add the last_digit to it)
repeat steps 2 to 4, till the original_number becomes 0. When original_number = 0, reversed_number would have the reverse of the original_number.

More info on step 4: If you are provided with a digit at a time, and asked to append it at the end of a number, how would you do it - by moving the original number one place to the left so as to accommodate the new digit. If number 23 has to become 234, you multiply 23 with 10 and then add 4.

234 = 23x10 + 4;

Code:

public static int reverseInt(int original_number) {
        int reversed_number = 0;
        while (original_number > 0) {
            int last_digit = original_number % 10;
            original_number = original_number / 10;
            reversed_number = reversed_number * 10 + last_digit;    
        }
        return reversed_number;
    }

User · Answer

while  input    0      reversedNum   reversedNum   10   input   10    input   input   10      let a number be 168     input   10 returns last digit as reminder i e  8 but next time it should return 6 hence number must be reduced to 16 from 168  as divide 168 by 10 that results to 16 instead of 16 8 as variable input is supposed to be integer type in the above program

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import java util Scanner   public class ReverseOfInteger       static Scanner input   new Scanner System in        public static void main String   args            int x   input nextInt            System out print helpermethod x               public static String helpermethod int x            if  x    0              return             String a   String valueOf x   10           return a   helpermethod x   10

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A method to get the greatest power of ten smaller or equal to an integer   in recursion   public static int powerOfTen int n        if   n  lt  10          return 1      else         return 10   powerOfTen n 10        The method to reverse the actual integer  in recursion   public static int reverseInteger int i        if  i   10  lt  1          return i       else         return i 10 powerOfTen i    reverseInteger i 10

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Simply you can use this      public int getReverseInt int value            int resultNumber   0          for  int i   value  i   0  i    10                resultNumber   resultNumber   10   i   10                    return resultNumber                  You can use this method with the given value which you want revers

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I am not clear about your Odd number  The way this code works is  it is not a Java specific algorithm  Eg  input  2345 first time in the while loop  rev 5 input 234 second time rev 5 10 4 54 input 23 third time rev 54 10 3 input 2 fourth time rev 543 10 2 input 0  So the reversed number is 5432  If you just want only the odd numbers in the reversed number then  The code is   while  input    0            last digit   input   10      if  last digit   2    0                 reversedNum   reversedNum   10   last digit             input   input   10

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while  num    0        rev   rev   10   num   10      num    10      That is the solution I used for this problem  and it works fine  More details   num   10   This statement will get you the last digit from the original number   num    10   This statement will eliminate the last digit from the original number  and hence we are sure that while loop will terminate   rev   rev   10   num   10   Here rev 10 will shift the value by left and then add the last digit from the original   If the original number was 1258  and in the middle of the run time we have rev   85  num   12 so   num 10   2  rev 10   850  rev 10   num 10   852

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public int getReverseNumber int number        int reminder   0  result   0      while  number   0                if  number  gt   10    number  lt   -10                        reminder   number   10              result   result   reminder              result   result   10              number   number   10                    else                       result   result   number              number    10                      return result          The above code will work for negative numbers also

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public static void reverse int number        while  number    0            int remainder   number   10          System out print remainder           number   number   10             System out println        What this does is  strip the last digit  within the 10s place  and add it to the front and then divides the number by 10  removing the last digit

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Even if negative integer is passed then it will give the negative integer Try This     public int reverse int result         long newNum 0 old result      result  result gt 0    result  0-result        while result  0           newNum  10          newNum  result 10          result  10          if newNum gt Integer MAX VALUE  newNum lt Integer MIN VALUE              return 0            if old  gt  0          return  int newNum      else if old  lt  0          return  int  newNum -1       else          return 0

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If you wanna reverse any number like 1234 and you want to revers this number to let it looks like 4321  First of all  initialize 3 variables int org   int reverse   0  and int reminder   then put your logic like       Scanner input   new Scanner  System in       System out println  Enter number to reverse         int org   input nextInt        int getReminder      int r   0      int count   0       while  org   0           getReminder   org 10           r   10   r   getReminder           org   org 10                   System out println r

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public static int reverseInt int i        int reservedInt   0       try          String s   String valueOf i           String reversed   reverseWithStringBuilder s           reservedInt   Integer parseInt reversed         catch  NumberFormatException e           System out println  exception caught was     e getMessage               return reservedInt     public static String reverseWithStringBuilder String str        System out println str       StringBuilder sb   new StringBuilder str       StringBuilder reversed   sb reverse        return reversed toString

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int aa 456  int rev Integer parseInt new StringBuilder aa     reverse

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int convert  int n            long val   0           if n  0              return 0           for int i   1  n  gt  exponent 10    i-1    i                          int mod   n    exponent 10  i                  int index   mod    exponent 10  i-1                 val    10              val    index                     if  val  lt  Integer MIN VALUE    val  gt  Integer MAX VALUE                         throw new IllegalArgumentException                  val     cannot be cast to int without changing its value                       return  int  val           static int exponent int m  int n                if n  lt  0               return 0          if 0    n               return 1           return  m   exponent m  n-1

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It is an outdated question  but as a reference for others First of all reversedNum must be initialized to 0   input 10 is used to get the last digit from input  input 10 is used to get rid of the last digit from input  which you have added to the reversedNum  Let s say input was 135   135   10 is 5 Since reversed number was initialized to 0 now reversedNum will be 5  Then we get rid of 5 by dividing 135 by 10  Now input will be just 13  Your code loops through these steps until all digits are added to the reversed number or in other words untill input becomes 0

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Reversing integer    int n  reverse   0    Scanner in   new Scanner System in     n   in nextInt       while n    0            reverse   reverse   10        reverse   reverse   n 10        n   n 10         System out println  Reverse of the number is     reverse

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Just to add on  in the hope to make the solution more complete   The logic by  sheki already gave the correct way of reversing an integer in Java  If you assume the input you use and the result you get always fall within the range  -2147483648  2147483647   you should be safe to use the codes by  sheki  Otherwise  it ll be a good practice to catch the exception    Java 8 introduced the methods addExact  subtractExact  multiplyExact and toIntExact  These methods will throw ArithmeticException upon overflow  Therefore  you can use the below implementation to implement a clean and a bit safer method to reverse an integer  Generally we can use the mentioned methods to do mathematical calculation and explicitly handle overflow issue  which is always recommended if there s a possibility of overflow in the actual usage   public int reverse int x        int result   0       while  x    0           try               result   Math multiplyExact result  10               result   Math addExact result  x   10               x    10            catch  ArithmeticException e                result   0     Exception handling             break                       return result

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I used String and I converted initially the int to String Then I used the reverse method  I found the reverse of the number in String and then I converted the string back to int  Here is the program   import java util     public class Panathinaikos       public static void my try                 Scanner input   new Scanner System in           System out println  Enter the number you want to be reversed            int number   input nextInt            String sReverse   Integer toString number           String reverse   new StringBuffer sReverse  reverse   toString            int Reversed   Integer parseInt reverse           System out print  The number     number    reversed is     Reversed

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Java solution without the loop  Faster response   int numberToReverse   your number  StringBuilder sb new StringBuilder    sb append numberToReverse   sb sb reverse    String intermediateString sb toString    int reversedNumber Integer parseInt intermediateString

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Scanner input   new Scanner System in           System out print  quot Enter number    quot            int num   input nextInt             System out print  quot Reverse number     quot            int value          while  num  gt  0               value   num   10              num      10              System out print value      value   Reverse

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import java util Scanner   public class Reverse order integer       private static Scanner scan       public static void main String   args            System out println   t t tEnter Number which you want to reverse  n            scan   new Scanner System in           int number   scan nextInt            int rev number   reverse number           System out println   t t tYour reverse Number is         rev number                                   n               private static int reverse int number            int backup   number          int count   0          while  number    0                number   number   10              count                      number   backup          int sum   0          for  int i   count  i  gt  0  i--                int sum10   1              int last   number   10              for  int j   1  j  lt  i  j                      sum10   sum10   10                            sum   sum    last   sum10               number   number   10                    return sum

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public static int reverse int x        boolean negetive   false      if  x  lt  0            x   Math abs x           negative   true             int y   0  i   0      while  x  gt  0            if  i  gt  0                y    10                     y    x   10          x   x   10          i              return negative   -y   y

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public static int reverse int x        int tmp   x      int oct   0      int res   0      while  true            oct   tmp   10          tmp   tmp   10          res    res oct  10          if   tmp 10     0                res   res tmp              return res

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Java reverse an int value - Principles   Modding     the input int by 10 will extract off the rightmost digit  example   1234   10    4 Multiplying an integer by 10 will  push it left  exposing a zero to the right of that number  example   5   10    50 Dividing an integer by 10 will remove the rightmost digit   75   10    7   Java reverse an int value - Pseudocode   a  Extract off the rightmost digit of your input number   1234   10    4  b  Take that digit  4  and add it into a new reversedNum   c  Multiply reversedNum by 10  4   10    40  this exposes a zero to the right of your  4    d  Divide the input by 10   removing the rightmost digit    1234   10    123  e  Repeat at step a with 123  Java reverse an int value - Working code  public int reverseInt int input        long reversedNum   0      long input long   input       while  input long    0            reversedNum   reversedNum   10   input long   10          input long   input long   10             if  reversedNum  gt  Integer MAX VALUE    reversedNum  lt  Integer MIN VALUE            throw new IllegalArgumentException              return  int  reversedNum      You will never do anything like this in the real work-world   However  the process by which you use to solve it without help is what separates people who can solve problems from the ones who want to  but can t unless they are spoon fed by nice people on the blogoblags

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See to get the last digit of any number we divide it by 10 so we either achieve zero or a digit which is placed on last and when we do this continuously we get the whole number as an integer reversed       int number 8989 last num sum 0      while number gt 0       last num number 10     this will give 8989 10 9     number  10         now we have 9 in last and now num  by 10  898     sum sum 10 last number      sum 0 10 9 9               last num 9    number  898  sum 9        last num 8    number  89   sum 9 10 8  98       last num 9    number 8     sum 98 10 9 989       last num 8    number 0     sum 989 10 8 9898      hence completed    System out println  Reverse is  sum

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This is the shortest code to reverse an integer  int i 5263   System out println Integer parseInt new StringBuffer String valueOf i    reverse   toString

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You can use recursion to solve this   first get the length of an integer number by using following recursive function   int Length int num int count       if num  0           return count            else          count            return Lenght num 10 count             and then you can simply multiply remainder of a number by 10  Length of integer - 1     int ReturnReverse int num int Length int reverse       if Length  0           reverse   reverse     num 10     int  Math pow 10 Length-1             return ReturnReverse num 10 Length-1 reverse             return reverse      The whole Source Code     import java util Scanner   public class ReverseNumbers        int Length int num  int count            if  num    0                return count            else               return Length num   10  count   1                        int ReturnReverse int num  int Length  int reverse            if  Length    0                reverse   reverse     num   10     int   Math pow 10  Length - 1                 return ReturnReverse num   10  Length - 1  reverse                     return reverse             public static void main String   args            Scanner scanner   new Scanner System in            int N   scanner nextInt             ReverseNumbers reverseNumbers   new ReverseNumbers            reverseNumbers ReturnReverse N  reverseNumbers Length N  0   reverseNumbers ReturnReverse N  reverseNumbers Length N  0   0             scanner close

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It s good that you wrote out your original code  I have another way to code this concept of reversing an integer  I m only going to allow up to 10 digits  However  I am going to make the assumption that the user will not enter a zero   if  inputNum  lt   999999999  amp  amp  inputNum  gt  0         System out print  Your number reversed is          do            endInt   inputNum   10    to get the last digit of the number       inputNum    10        system out print endInt           While inputNum    0    System out println          else    System out println  You used an incorrect number of integers  n     System out println  Program end

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public static double reverse int num        double num1   num      double ret   0      double counter   0       while  num1  gt  1                   counter            num1   num1 10            while counter  gt   0                int lastdigit   num 10          ret    Math pow 10  counter-1    lastdigit          num   num 10          counter--              return ret

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public static void main String args          int n   0  res   0  n1   0  rev   0      int sum   0      Scanner scan   new Scanner System in       System out println  Please Enter No           n1   scan nextInt       String s1 String valueOf n1       int len    n1    0    1    int  Math log10 n1    1      while  n1  gt  0            rev   res     int  Math pow 10  len            res   n1   10          n1   n1   10             sum  res    sum sum res          sum    rev          len--               System out println  sum No      sum       System out println  sum No       sum   res        This will return reverse of integer

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123 maps to 321  which can be calculated as 3  10 2  2  10 1  1 Two functions are used to calculate  10 N   The first function calculates the value of N  The second function calculates the value for ten to power N   Function lt Integer  Integer gt  powerN   x - gt  Double valueOf Math log10 x   intValue    Function lt Integer  Integer gt  ten2powerN   y - gt  Double valueOf Math pow 10  y   intValue        123   gt  321  3 10 2   2 10   1 public int reverse int number        if  number  lt  10            return number        else           return  number   10    powerN andThen ten2powerN  apply number    reverse number   10

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import java io BufferedReader  import java io InputStreamReader  public class intreverse   public static void main String   a throws Exception       int no      int rev   0      System out println  Enter The no to be reversed        InputStreamReader str new InputStreamReader System in       BufferedReader br  new BufferedReader str       no Integer parseInt br readLine   toString         while no  0                rev rev 10 no 10          no no 10             System out println rev

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Here is a complete solution returns 0 if number is overflown    public int reverse int x        boolean flag   false          Helpful to check if int is within range of  int      long num   x          if the number is negative then turn the flag on      if x  lt  0            flag   true          num   0 - num                used for the result      long result   0          continue dividing till number is greater than 0     while num  gt  0            result   result 10   num 10          num  num 10             if flag            result   0 - result             if result  gt  Integer MAX VALUE    result  lt  Integer MIN VALUE            return 0            return  int  result

[java] Java reverse an int value without using array

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