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(easy)LeetCode 258.Add Digits

Posted on 2015-08-19 16:31  骄阳照林  阅读(122)  评论(0编辑  收藏  举报

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 = 111 + 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;
    }
}