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?

解法

先上代码：

public int addDigits(int num) { return num == 0 ? 0 : (num%9 == 0 ? 9 : num%9); }

一个数的树根：

证明： wiki：https://en.wikipedia.org/wiki/Digital_root 知乎：https://www.zhihu.com/question/30972581