1. 问题描述

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:

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

Tags: Math

Similar Problems: (E) Happy Number

2. 解题思路

 不用 循环 与 递归，毫无思路，从最简单情况开始分析：

• 对于一位数：0、1、2、3、4、5、6、7、8、9时，返回该数即可
• 对于两位数：10、11、12、13...，可发现规律，该返回 1 + (num-1)%9
• 对于三位数：100、101、102、103...，该规律同样适用，该返回 1 + (num-1)%9

3. 代码

            1
            class
             Solution {

            2
            public
            :

            3
            int addDigits(int num) 
4    {
5return1 + (num-1)%9; 
6    }
7 }

4. 反思

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

原文：http://www.cnblogs.com/whl2012/p/5596675.html