15算法课程 258. Add Digits



Given a non-negative integer num, repeatedly add all its digits until the result has only one digit.

For example:

Given num = 38, the process is like: 3 + 8 = 11, 1 + 1 = 2. Since 2 has only one digit, return it.

Follow up:
Could you do it without any loop/recursion in O(1) runtime?

solution:

递归不断求和

code:

class Solution {public:    int addDigits(int num) {        if(num<10)            return num;        else            return addDigits(sum(num));    }    int sum(int num){        int sum = 0;        while(num)        {            sum+=num%10;            num = num/10;        }        return sum;    }};