Sum of Digits in an Integer

Given a non-negative int n, return the sum of its digits recursively (no loops). Note that mod (%) by 10 yields the rightmost digit (126 % 10 is 6), while divide (/) by 10 removes the rightmost digit (126 / 10 is 12).



sumDigits(126) → 9


sumDigits(49) → 13


sumDigits(12) → 3
----------------------------------------------------------
public int sumDigits(int n) {




int Count=0;

if(n/10<10){

Count=(n%10)+(n/10);

}

else{

Count=sumDigits(n/10)+sumDigits(n%10);

}

return Count;



}

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