leetcode--Add Digits

题目：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?

One：基本循环 ，先求各数之和，后把和赋给num，直到num<=9

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

Two:不用循环，注意：num的返回值是除以9的余数。

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