Add Digits

https://leetcode.com/problems/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?

1 1
2 2
3 3
4 4
5 5
6 6
7 7
8 8
9 9
10 1
11 2
12 3
13 4
14 5
15 6
16 7
17 8
18 9
19 1
20 2

根据上面的列举，我们可以得出规律，每9个一循环，所有大于9的数的树根都是对9取余，那么对于等于9的数对9取余就是0了，为了得到其本身，而且同样也要对大于9的数适用，我们就用(n-1)%9+1这个表达式来包括所有的情况，所以解法如下：

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

参考
http://www.cnblogs.com/grandyang/p/4741028.html