[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;    }};