Leet Code OJ 258. Add Digits [Difficulty: Easy]

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?

代码实现：

public class Solution {
    public int addDigits(int num) {
        if(num>9){
            return addDigits(num%10+addDigits(num/10));
        }else{
            return num;
        }
    }
}