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?

class Solution {
public:
    int addDigits(int num) {
        if( num == 0)
            return 0;
        else
            return ( num - 1 ) % 9 + 1;
    }
};

找规律，发现结果是模9循环的。数字类型的题目，很多都是找规律。