Leetcode 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?

将一个数的各个位加起来，如此反复知道只剩各位数。不用循环和递归在常数时间复杂度内解决。

列几个数就可以发现规律，除了0以外其他数以1-9循环。

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