LeetCode 258. Add Digits

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：
其实就是如果这个数是0，返回0；这个数是9的倍数，直接返回9；若不是9的倍数，则返回它除9的余数。

代码：

class Solution {
public:int addDigits(int num) {if(num==0){return 0;}else if(num%9==0){return 9;}else{return num%9;}}
};

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