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?

 

这题主要是多写出许多数据,然后找出规律。

 

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

 

posted on 2017-12-16 21:15  夜的第八章  阅读(126)  评论(0编辑  收藏  举报

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