Sum of all digits for a given Positive Number [closed]

Question

Method return should be like if entered a number, suppose 345, then output should be 3+4+5=12 --> 1+2 = 3. what i am doing wrong here?

public class DigitSum
 {
    int  Sum=0;

    public int compute( int MethParam )
    {
        int rem = MethParam%10; 
        Sum+=rem;        

        MethParam = MethParam/10; 
        if(MethParam>10)
            compute(MethParam);

        return Sum+MethParam;  
    }

  public static void main(String[] args)
  {
    DigitSum ds  = new DigitSum();
    System.out.println(ds.compute(435));
  }
}

Answer 1

O(1) Algo for sum Of digits :

Taking the modulo 9 of any number will return the sum of digits of that number until a single digit number is obtained.

If the number is a multiple of 9, then the sum will will be 9

one liner :

public int sumDigit(int n){
    return (n%9 == 0 && n != 0) ? 9 : n%9;
}

Alternate implementation :

public int sumDigit(int n){

      int sum = n % 9;
      if(sum == 0){
          if(n > 0)
               return 9;
      }
      return sum;
}

Answer 2

What you are looking for is digital root. So here's a better solution using the formula from wiki page I linked.

Without Recursion: -

public static int compute( int n ) {
    return n - 9 * ((n - 1) / 9);
}

---

And, just in case you want (Which I don't think you would), here's a one-liner (Using Recursion): -

public static int compute( int n ) {
    return n < 10 ? n : compute(n % 10 + compute(n / 10));
}

Answer 3

    public int FindSumDigit(int number)
    {
        if (number < 10) return number;
        int sum = 0;
        while (number > 0)
        {
            sum += number % 10;
            number = number / 10;
        }
        return FindSumDigit(sum);
    }

Find my code... Poon you were not adding the whole digits.. In middle itself u was keep on adding the right most digit.

Answer 4

Many wrong answers here. Here's what OP wants:

Method return should be like if entered a number, suppose 345, then output should be 3+4+5=12 --> 1+2 = 3.

This will do the job:

public static int compute(int param) {
    int sum = 0;
    do {
        sum = 0;
        while (param > 0) {
            sum += param % 10;
            param /= 10;
        }
        param = sum;
    } while (sum >= 10);
    return sum;
}

Answer 5

I changed your method to this, then it gives the requested result:

public int compute(int methParam) {
    int sum = 0;
    for (int i = 0; methParam > 10; i++) {
        int currentDigit = methParam % 10;
        methParam = methParam / 10;
        sum = sum + currentDigit;
    }
    if (sum + methParam > 10) {
        return compute(sum + methParam);
    } else {
        return sum + methParam;
    }
}

Please note that i moved the declaration of sum inside the method, instead of making it a field.

Answer 6

IN your code your are not properly returning values to get call for your recursion method.

        if ((MethParam >= 10)){
            return compute(MethParam);
        }else
            return Sum + MethParam;