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

Answer 1

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 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

try

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);
    else
        Sum+=MethParam;

        if(Sum>=10){
        int temp=Sum;
        Sum=0;
        compute(temp);
    }

         return Sum;  
}

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

Answer 4

just a different approach, may not be as efficient due to involved conversions:

private static int getUltimateSum(int input) {
    String inputStr = String.valueOf(input);
    int sum = 0;
    for (int i = 0; i < inputStr.length(); i++) {
        int digit = Integer.valueOf(String.valueOf(inputStr.charAt(i)));
        sum += digit;
    }
    if(sum > 9)
        return getUltimateSum(sum);
    else
        return sum;
}

Answer 5

use this simple java code I created some tymz ago, this was for adding positive numbers as well as negative numbers:

class SumDigit
{
public static void main(String args[])
{
int sum, i,a,d;
a = Integer.parseInt(args[0]);
sum = 0;
for(i=1;i< =10;i++)
{
d = a%10;
a = a/10;
sum=sum + d;
} 
System.out.println("Sum of Digit :"+sum);
}
}

Answer 6

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;
}