258. Add Digits

1.描述

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?

2.分析

3.代码

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