258. 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?

这个很简单了。与9同余，关键就是如果n不是0的话得到的余数不能为0而是为9

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