leetcode之旅(6)-Add Digits

题目:

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?

Show Hint 

题目分析:

首先觉得例子很明显,可能的结果是数目固定的(0~9),那么很有可能是由规律的,加上后面的一句,要求在常数范围内做出来,那么很可能有很大的规律,从0~30总结 一下规律

总结得出结论是从1~9循环

我在审题时候,遗漏了非负数,包括0,没有考虑0的特殊情况

代码:

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

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