How to sum digits of an integer in java

Question

I am having a hard time figuring out the solution to this problem  I am trying to develop a program in Java that takes a number  such as 321  and finds the sum of digits  in this case 3   2   1   6  I need all the digits of any three digit number to add them together  and store that value using the   remainder symbol  This has been confusing me and I would appreciate anyones ideas

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May be little late   but here is how you can do it recursively  public int sumAllDigits int number        int sum   number   10       if number 10  lt  10           return sum   number 10       else          return sum   sumAllDigits number 10

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In Java 8   public int sum int number      return  number       chars                          map digit - gt  digit   48                         sum        Converts the number to a string and then each character is mapped to it s digit value by subtracting ascii value of  0   48  and added to the final sum

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public static void main String   args            int num   321          int sum   0          while  num  gt  0                sum   sum   num   10              num   num   10                    System out println sum        Output  6

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If you love constant time try this   double d   10984 491      converting to String because of floating point issue of precision String s   new String d       replaceAll    D        int i   Integer parseInt s          System out println i   9    0   9   i   9     Logic is if you add any number by 9  resulting addition of digit will result in the same number       Example  6   9   15 then 1   5   6  again you got 6      In case of decimal point  remove it and add resulting digits   Below code does the trick   i   9    0   9   i   9

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Java 8 Recursive Solution  If you dont want to use any streams   UnaryOperator lt Long gt  sumDigit   num - gt  num  lt   0   0   num   10   this sumDigit apply num 10     How to use  Long sum   sumDigit apply 123L     Above solution will work for all positive number  If you want the sum of digits irrespective of positive or negative also then use the below solution   UnaryOperator lt Long gt  sumDigit   num - gt  num  lt   0              num    0   0   this sumDigit apply -1   num               num   10   this sumDigit apply num 10

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Click here to see full program  Sample code    public static void main String args          int number   333      int sum   0      int num   number      while  num  gt  0            int lastDigit   num   10          sum    lastDigit          num    10            System out println  Sum of digits     sum

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without mapping   the quicker lambda solution  Integer toString  num   chars   boxed   collect  Collectors summingInt   c  - gt  c -  0            or same with the slower   operator  Integer toString  num   chars   boxed   collect  Collectors summingInt   c  - gt  c    0            or Unicode compliant  Integer toString  num   codePoints   boxed   collect  Collectors summingInt  Character  getNumericValue

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The following method will do the task   public static int sumOfDigits int n        String digits   new Integer n  toString        int sum   0      for  char c  digits toCharArray            sum    c -  0       return sum      You can use it like this   System out printf  Sum of digits    d n   sumOfDigits 321

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Java 8 Solution  int n  29      String valueOf n  chars   map Character  getNumericValue  sum

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If you need a one-liner I guess this is a very nice solution   int sum int n     return n  gt   10   n   10   sum n   10    n

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Recursions are always faster than loops   Shortest and best   public static long sumDigits long i        return i    0   0   i   10   sumDigits i   10

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Mine is more simple than the others hopefully you can understand this if you are a some what new programmer like myself   import java util Scanner  import java lang Math   public class DigitsSum        public static void main String   args             Scanner in   new Scanner System in           int digit   0          System out print  Please enter a positive integer              digit   in nextInt            int D1   0          int D2   0          int D3   0          int G2   0          D1   digit   100          D2   digit   100          G2   D2   10          D3   digit   10            System out println D3   G2   D1

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You can do it using Recursion    Sum of digits till single digit is obtained public int sumOfDigits int num                         int sum   0               while  num  gt  0                                sum   sum   num   10                  num   num   10                             sum    sum  lt 10    sum   sumOfDigits sum                return sum

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In addition to the answers here  I can explain a little bit  It is actually a mathematical problem   321 is the total of 300   20   1    If you divide 300 by 100  you get 3    If you divide 20 by 10  you get 2   If you divide 1 by 1  you get 1   At the end of these operations  you can sum up all of them and you get 6   public class SumOfDigits   public static void main String   args        int myVariable   542      int checker   1      int result   0      int updater   0         This while finds the size of the myVariable     while  myVariable   checker    myVariable            checker   checker   10               This for statement calculates  what you want      for  int i   checker   10  i  gt  0  i   i   10             updater   myVariable   i          result    updater          myVariable   myVariable -  updater   i              System out println  The result is     result

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Here is a simple program for sum of digits of the number 321           import java math             class SumOfDigits               public static void main String args    throws Exception                   int sum   0                  int i   321                      sum    i   10     i   10                           if  sum  gt  9                                int n    sum   10     sum   10                               System out print  Sum of digits of     i     is     n                             else                             System out print  Sum of digits of     i     is     sum                                               Output   Sum of digits of 321 is 6   Or simple you can use this  check below program   public class SumOfDigits    public static void main String   args         long num   321       int rem sum 0      while num  0                rem   num 10          sum   sum rem          num num 10            System out println sum               if num  0                long sum     num 9  0    9   num 9           System out println sum

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This should be working fine for any number of digits and it will return individual digit s sum  public static void main String   args        Scanner input   new Scanner System in       System out println  enter a string        String numbers   input nextLine       String would be 55     int sum   0      for  char c   numbers toCharArray              sum    c -  0             System out println sum     the answer is 10

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It might be too late  but I see that many solutions posted here use O n 2  time complexity  this is okay for small inputs  but as you go ahead with large inputs  you might want to reduce time complexity  Here is something which I worked on to do the same in linear time complexity   NOTE   The second solution posted by Arunkumar is constant time complexity       private int getDigits int num        int sum  0      while num  gt  0      num consists of 2 digits max  hence O 1  operation         sum   sum   num   10          num   num   10               return sum    public int addDigits int N        int temp1 0  temp2  0      while N  gt  0            temp1  N   10          temp2  temp1   temp2          temp2  getDigits temp2      this is O 1  operation         N   N  10            return temp2         Please ignore my variable naming convention  I know it is not ideal  Let me explain the code with sample input   e g   12345   Output must be 6  in a single traversal   Basically what I am doing is that I go from LSB to MSB   and add digits of the sum found  in every iteration    The values look like this  Initially temp1   temp2   0  N       temp1   N   10     temp2   temp1   temp2   12345   5                  5    1234    4                  5   4   9   getDigits 9    9  123     3                  9   3   12   3  getDigits 12   3   12      2                  3   2   5  getDigits 5    5  1       1                  5   1   6  getDigits 6    6     Answer is 6  and we avoided one extra loop  I hope it helps

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A simple solution using streams   int n   321  int sum   String valueOf n       chars        map Character  getNumericValue       sum

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shouldn t you be able to do it recursively something like so  I m kinda new to programming but I traced this out and I think it works   int sum int n   return n 10   sum n 10

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Sums all the digits regardless of number s size      private static int sumOfAll int num        int sum   0       if num    10  return 1       if num  gt  10            sum    num   10          while  num   num   10   gt   1                sum     num  gt  10    num 10   num                  else           sum    num            return sum

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In Java 8  this is possible in a single line of code as follows   int sum   Pattern compile              splitAsStream factorialNumber toString             mapToInt Integer  valueOf           sum

[java] How to sum digits of an integer in java?

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