[Leetcode] 258. Add Digits 解题报告

题目：

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?

思路：

没什么可说的。

代码：

class Solution {
public:
    int addDigits(int num) {
        int sum = 0;
        while (num > 9) {
    		sum = 0;
    		while (num != 0) {
    			sum += num % 10;
    			num /= 10;
    		}
    		num = sum;
    	}
    	return num;
    }
};