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?

s思路：
1. 如何下手？不用loop.也就是说不能一位一位的取，不用考虑细节的来做，方法就两种，不是bottom-up,就是top-down.
2. 这就是说用一个top-down的系统方法，一般数学的方法就比较好：参考了https://en.wikipedia.org/wiki/Digital_root
3. 就是n%9的余数，如果余数等于0，就返回9.所以修正一下就是:(n-1)%9+1
4. 应该能猜测到，最后结果是0-9，所以才都可以猜一下用余数

class Solution {
public:
    int addDigits(int num) {
        if(num<1) return num;
        return (num-1)%9+1;
    }
};