(easy)LeetCode 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?

方法1：常规做法，循环，不符合题意。

       

public class Solution {
    public int addDigits(int num) {
        while(num>9){
            int p=0;
             while(num!=0){
                 p+=num%10;
                 num=num/10;
             }
             num=p;
        }
        return num;
    }
}

方法2：找到规律，一行代码

 代码如下：

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