LeetCode - Add Digits

题目原址：https://leetcode.com/problems/add-digits/?tab=Description

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?

这道题的关键在于题目要求时间复杂度为O(1), 显然一般的方法是做不到的，这里我们用 repeated digital sum，也叫 digital root 算法直接求解。

Congruence formula： f(num)=1+(num−1)%9

关于这个算法的详细介绍，大家可以看wiki，讲的非常清楚： https://en.wikipedia.org/wiki/Digital_root

最后附上代码片段：

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