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?

//题目要求不能用递归、循环
//后来参考 优快云 博主 nomasp 的贴出的代码写的。
//公式：dr(n) = 1 + ((n-1) mod 9).
class Solution {
public:
    int addDigits(int num) {
        return 1 + (num-1) % 9;
    }
};

附上传送门：
http://blog.youkuaiyun.com/nomasp/article/details/49764711