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leetCode(62)-Reverse Integer

时间:2016-04-16 19:23:18      阅读:140      评论:0      收藏:0      [点我收藏+]

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题目:

Reverse digits of an integer.

Example1: x = 123, return 321
Example2: x = -123, return -321

click to show spoilers.

Have you thought about this?
Here are some good questions to ask before coding. Bonus points for you if you have already thought through this!

If the integer‘s last digit is 0, what should the output be? ie, cases such as 10, 100.

Did you notice that the reversed integer might overflow? Assume the input is a 32-bit integer, then the reverse of 1000000003 overflows. How should you handle such cases?

For the purpose of this problem, assume that your function returns 0 when the reversed integer overflows.

Update (2014-11-10):
Test cases had been added to test the overflow behavior.

思路:

  • 题意:反转整数
  • 注意两点:1.负数的情况2.溢出返回0(绝对值大于Integer.MAX_VALUE)

代码:

public class Solution {
    public int reverse(int x) {
        boolean flag = true;
        if(x < 0){
            flag = false;
            x = x*-1;
        }
        long res = 0;
        while(x > 0){
            res = res *10 + x%10;
            x = x / 10;
        }
        if(res > Integer.MAX_VALUE){
            return 0;
        }
        if(flag){
            return (int)res;
        }else{
            return (int)res * -1;
        }
    }
}

leetCode(62)-Reverse Integer

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原文地址:http://blog.csdn.net/lpjishu/article/details/51161042

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