258. Add Digits

题目：

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?

思路：

乍一看确实没有思路，但是后来在看了discuss之后明白了这又是一个找规律的问题。

代码如下：

public class Solution {
public int addDigits(int num) {
if(num == 0){
return 0;
}
int mynum = num % 9;
if(mynum == 0){
mynum = 9;
}
return mynum;
}
}

需要注意的是需要将模数得到的0值改成9.