(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;
}
}