How to get the separate digits of an int number

Question

I have numbers like 1100  1002  1022 etc  I would like to have the individual digits  for example for the first number 1100 I want to have 1  1  0  0   How can I get it in Java

User · Answer

in Java  this is how you can separate digits from numbers and store them in an Array   public static void main String   args            System out println  Digits Array     Arrays toString getNumberArr 1100        private static Integer   getNumberArr int number          will get the total number of digits in the number     int temp   number      int counter   0       while  temp  gt  0            temp    10          counter                reset the temp     temp   number          make an array     int modulo        modulo is equivalent to single digit of the number      Integer   numberArr   new Integer counter       for  int i   counter - 1  i  gt   0  i--            modulo   temp   10          numberArr i    modulo            temp    10             return numberArr      Output   Digits Array    1  1  0  0

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Here is my answer  I did it for myself and I hope it s simple enough for those who don t want to use the String approach or need a more math-y solution   public static void reverseNumber2 int number         int residual 0      residual number 10      System out println residual        while  residual  number               number  number-residual  10            residual number 10            System out println residual             So I just get the units  print them out  substract them from the number  then divide that number by 10 - which is always without any floating stuff  since units are gone  repeat

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Try this   int num  4321 int first     num   10  int second      num - first     100   10  int third       num - first - second     1000   100  int fourth      num - first - second - third     10000   1000    You will get first   1  second   2  third   3 and fourth    4

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int number   12344444     or it Could be any valid number  int temp   0  int divider   1   for int i  1  i lt  String valueOf number  length   i            divider   divider   10      while  divider  gt 0         temp   number   divider      number   number   divider      System out print temp            divider   divider 10

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neither chars   nor codePoints       the other lambda  String number   Integer toString  1100     IntStream range  0  number length     map  i - gt  Character digit  number codePointAt  i    10     toArray         1  1  0  0

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I noticed that there are few example of using Java 8 stream to solve your problem but I think that this is the simplest one   int   intTab   String valueOf number  chars   map Character  getNumericValue  toArray      To be clear  You use String valueOf number  to convert int to String  then chars   method to get an IntStream  each char from your string is now an Ascii number   then you need to run map   method to get a numeric values of the Ascii number  At the end you use toArray   method to change your stream into an int   array

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if digit is meant to be a Character String numstr   Integer toString  123    Pattern compile   quot  quot    splitAsStream  numstr   map    s - gt  s charAt  0     toArray  Character    new         1  2  3   and the following works correctly numstr    quot 000123 quot  gets  0  0  0  1  2  3  numstr    quot -123 quot       gets  -  1  2  3

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Something like this will return the char     public static char   getTheDigits int value       String str           int number   value      int digit   0      while number gt 0           digit   number 10          str   str   digit          System out println  Digit     digit           number   number 10                  return str toCharArray

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I haven t seen anybody use this method  but it worked for me and is short and sweet    int num   5542  String number   String valueOf num   for int i   0  i  lt  number length    i          int j   Character digit number charAt i   10       System out println  digit      j       This will output    digit  5 digit  5 digit  4 digit  2

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How about this   public static void printDigits int num        if num   10  gt  0            printDigits num   10             System out printf   d    num   10       or instead of printing to the console  we can collect it in an array of integers and then print the array   public static void main String   args        Integer   digits   getDigits 12345       System out println Arrays toString digits       public static Integer   getDigits int num        List lt Integer gt  digits   new ArrayList lt Integer gt         collectDigits num  digits       return digits toArray new Integer          private static void collectDigits int num  List lt Integer gt  digits        if num   10  gt  0            collectDigits num   10  digits             digits add num   10       If you would like to maintain the order of the digits from least significant  index 0   to most significant  index n    the following updated getDigits   is what you need          split an integer into its individual digits    NOTE  digits order is maintained - i e  Least significant digit is at index 0      param num positive integer     return array of digits     public static Integer   getDigits int num        if  num  lt  0    return new Integer 0         List lt Integer gt  digits   new ArrayList lt Integer gt         collectDigits num  digits       Collections reverse digits       return digits toArray new Integer

User · Answer

This uses the modulo 10 method to figure out each digit in a number greater than 0  then this will reverse the order of the array  This is assuming you are not using  0  as a starting digit   This is modified to take in user input  This array is originally inserted backwards  so I had to use the Collections reverse   call to put it back into the user s order       Scanner scanNumber   new Scanner System in       int userNum   scanNumber nextInt       user s number         divides each digit into its own element within an array     List lt Integer gt  checkUserNum   new ArrayList lt Integer gt         while userNum  gt  0            checkUserNum add userNum   10           userNum    10             Collections reverse checkUserNum      reverses the order of the array      System out print checkUserNum

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Integer toString 1100  gives you the integer as a string   Integer toString 1100  getBytes   to get an array of bytes of the individual digits   Edit   You can convert the character digits into numeric digits  thus     String string   Integer toString 1234     int   digits   new int string length        for int i   0  i lt string length      i       digits i    Integer parseInt string substring i  i 1          System out println  digits     Arrays toString digits

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As a noob  my answer would be   String number   String valueOf ScannerObjectName nextInt      int   digits   new int number length      for  int i   0   i  lt  number length     i        int i    Integer parseInt digits substring i i 1     Now all the digits are contained in the  digits  array

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Since I don t see a method on this question which uses Java 8  I ll throw this in  Assuming that you re starting with a String and want to get a List lt Integer gt   then you can stream the elements like so  List lt Integer gt  digits   digitsInString chars            map Character  getNumericValue           boxed            collect Collectors toList      This gets the characters in the String as a IntStream  maps those integer representations of characters to a numeric value  boxes them  and then collects them into a list

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Easier way I think is to convert the number to string and use substring to extract and then convert to integer   Something like this   int digits1  Integer parseInt  String valueOf 201432014  substring 0 4        System out println  digits are    digits1     ouput is  2014

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I see all the answer are ugly and not very clean   I suggest you use a little bit of recursion to solve your problem  This post is very old  but it might be helpful to future coders   public static void recursion int number        if number  gt  0            recursion number 10           System out printf   d       number 10              Output   Input  12345  Output  1   2   3   4   5

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I wrote a program that demonstrates how to separate the digits of an integer using a  more simple and understandable approach that does not involve arrays  recursions  and all that fancy schmancy  Here is my code   int year   sc nextInt    temp   year  count   0   while  temp gt 0      count      temp   temp   10     double num   Math pow 10  count-1   int i    int num   for   i gt 0 i  10      System out println year i 10       Suppose your input is the integer 123  the resulting output will be as follows   1 2 3

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Java 9 introduced a new Stream iterate method which can be used to generate a stream and stop at a certain condition  This can be used to get all the digits in the number  using the modulo approach   int   a   IntStream iterate 123400  i - gt  i  gt  0  i - gt  i   10  map i - gt  i   10  toArray      Note that this will get the digits in reverse order  but that can be solved either by looping through the array backwards  sadly reversing an array is not that simple   or by creating another stream   int   b   IntStream iterate a length - 1  i - gt  i  gt   0  i - gt  i - 1  map i - gt  a i   toArray      or  int   b   IntStream rangeClosed 1  a length  map i - gt  a a length - i   toArray      As an example  this code   int   a   IntStream iterate 123400  i - gt  i  gt  0  i - gt  i   10  map i - gt  i   10  toArray    int   b   IntStream iterate a length - 1  i - gt  i  gt   0  i - gt  i - 1  map i - gt  a i   toArray    System out println Arrays toString a    System out println Arrays toString b      Will print    0  0  4  3  2  1   1  2  3  4  0  0

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Why don t you do  String number   String valueOf input   char   digits   number toCharArray

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simple solution   public static void main String   args        int v   12345      while  v  gt  0           System out println v   10           v    10

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could be any num this is a randomly generated one int num    int   Math random     1000       this will return each number to a int variable int num1   num   10  int num2   num   10   10  int num3   num  100   10      you could continue this pattern for 4 5 6 digit numbers    dont need to print you could then use the new int values man other ways System out print num1   System out print   n    num2   System out print   n    num3

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Convert it to String and use String toCharArray   or String split     String number   String valueOf someInt    char   digits1   number toCharArray       or  String   digits2   number split     lt          In case you re already on Java 8 and you happen to want to do some aggregate operations on it afterwards  consider using String chars   to get an IntStream out of it   IntStream chars   number chars

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public int   getDigitsOfANumber int number        String numStr   String valueOf number       int retArr     new int numStr length          for  int i   0  i  lt  numStr length    i              char c   numStr charAt i           int digit   c          int zero    char   0           retArr i    digit - zero             return retArr

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see bellow my proposal with comments            int size i toString   length       the length of the integer  i  we need to split             ArrayList lt Integer gt  li   new ArrayList lt Integer gt        an ArrayList in whcih to store the resulting digits          Boolean b true     control variable for the loop in which we will reatrive step by step the digits         String number  1      here we will add the leading zero depending on the size of i         int temp      the resulting digit will be kept by this temp variable      for  int j 0  j lt size  j                             number number concat  0                           Integer multi   Integer valueOf number      the variable used for dividing step by step the number we received                  while b                        multi multi 10                      temp i  multi                       li add temp                       i i  multi                                           if i  0                                           b false                                                                                 for Integer in  li                       System out print in intValue

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import java util Scanner   class  Test          public static void main String   args                     Scanner sc   new Scanner System in          int num sc nextInt         System out println  Enter a number  -1 to end    num       int result 0      int i 0      while true                int n num 10        if n  -1           break                i          System out println  Digit  i       n         result result 10 n        num num 10           if num  0                    break

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Java 8 solution to get digits as int   from an integer that you have as a String   int   digits   intAsString chars   map i - gt  i -  0   toArray

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import java util Scanner   public class SeparatingDigits        public static void main  String   args                  System out print   Enter the digit to print separately  -             Scanner scan   new Scanner  System in             int element1   scan nextInt            int divider           if    element1  gt  9999    amp  amp    element1  lt   99999                           divider   10000                    else if    element1  gt  999    amp  amp    element1  lt   9999                           divider   1000                    else if     element1  gt  99   amp  amp    element1  lt   999                           divider   100                    else if    element1  gt  9    amp  amp    element1  lt   99                           divider   10                    else                        divider   1                     quotientFinder  element1  divider                    public static void quotientFinder  int elementValue  int dividerValue                   for  int count   1   dividerValue    0  count                           int quotientValue   elementValue   dividerValue               elementValue   elementValue   dividerValue               System out printf    d     quotientValue                 dividerValue    10                             Without using arrays and Strings     digits 1-99999    output    Enter the digit to print separately  - 12345  1  2  3  4  5

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To do this  you will use the    mod  operator   int number       some int  while  number  gt  0        print  number   10       number   number   10      The mod operator will give you the remainder of doing int division on a number   So   10012   10   2   Because   10012   10   1001  remainder 2   Note  As Paul noted  this will give you the numbers in reverse order   You will need to push them onto a stack and pop them off in reverse order   Code to print the numbers in the correct order   int number       and int LinkedList lt Integer gt  stack   new LinkedList lt Integer gt     while  number  gt  0        stack push  number   10        number   number   10     while   stack isEmpty          print stack pop

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Just to build on the subject  here s how to confirm that the number is a palindromic integer in Java   public static boolean isPalindrome int input    List lt Integer gt  intArr   new ArrayList    int procInt   input   int i   0  while procInt  gt  0        intArr add procInt 10       procInt   procInt 10      i       int y   0  int tmp   0  int count   0  for int j intArr        if j    0  amp  amp  count    0        break             tmp   j    tmp 10       count       if input    tmp      return false   return true      I m sure I can simplify this algo further  Yet  this is where I am  And it has worked under all of my test cases   I hope this helps someone

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I think this will be the most useful way to get digits   public int   getDigitsOf int num                int digitCount   Integer toString num  length         if  num  lt  0           digitCount--                  int   result   new int digitCount        while  digitCount--  gt 0            result digitCount    num   10          num    10                    return result      Then you can get digits in a simple way   int number   12345  int   digits   getDigitsOf number    for  int i   0  i  lt  digits length  i          System out println digits i        or more simply   int number   12345  for  int i   0  i  lt  getDigitsOf number  length  i          System out println   getDigitsOf number  i          Notice the last method calls getDigitsOf method too much time  So it will be slower  You should create an int array and then call the getDigitsOf method once  just like in second code block   In the following code  you can reverse to process  This code puts all digits together to make the number   public int digitsToInt int   digits        int digitCount   digits length      int result   0       for  int i   0  i  lt  digitCount  i              result   result   10          result    digits i              return result      Both methods I have provided works for negative numbers too

[java] How to get the separate digits of an int number?

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