(easy) Leetcode 258.ADD Digits

Given a non-negative integer num , repeatedly add all its digits until the result have only one digit.

For example:

Given num = 38 , the process is like: 3 + 8 = 11 , 1 + 1 = 2 . Since have only one 2 digit, return it.

Follow up:
Could do it without any loop/recursion in O (1) runtime?

Method 1: General practice, cyclic, does not conform to test instructions.

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;    }}

Method 2: Find the rule, one line of code

The code is as follows:

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

　　

(easy) Leetcode 258.ADD Digits