258.LeetCode Add Digits(easy)[数学问题 求一个数的树根]

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?

Hint:

1. A naive implementation of the above process is trivial. Could you come up with other methods?
2. What are all the possible results?
3. How do they occur, periodically or randomly?
4. You may find this Wikipedia article useful.

参考维基百科上的https://en.wikipedia.org/wiki/Digital_root

这是一个求一个数的树根的问题，通过观察找规律可以发现一个数的树根就是这个数(num-1)%9+1的结果

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