258. Add Digits【规律】

2017/3/16 22:36:02

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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?

 

版本1    循环

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

　　

 

版本2   列出表格观察，发现9个数字一循环的规律，一行代码就搞定。

 

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