LeetCode 258. Add Digits(1+(n-1)%9)

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：ans(n) = 1 + (n - 1) % 9

Code:

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