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?

可能是直觉，瞬间发现这个规律，等到提完数据再review

class Solution {
public:
    int addDigits(int num) {
        if(!(num % 9) && num)
            return 9;
        return num % 9;
    }
};