LeetCode 7 Reverse Integer

最新推荐文章于 2025-09-12 17:31:31 发布

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原文链接：http://www.cnblogs.com/wood-python/p/5828458.html

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#数据结构与算法

本文介绍了一种用于反转整数的有效算法，并特别强调了处理负数和溢出情况的方法。通过使用字符串缓冲区来反转数字，确保了在面对特殊输入如边界值时能够正确返回结果。

Reverse digits of an integer.

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

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：

题目本身并不难，循环、递归、字符串类等方法均可解。这道题的坑在于虽然函数签名传入的参数是int类型，但是LeetCode上的测试用例远远超出int类型的范围，因此要注意溢出的情况。

解法：

 1 public class Solution
 2 {
 3     public int reverse(int x)
 4     {
 5         long number = x;
 6         int sign = 1;
 7 
 8         if(number < 0)
 9         {
10             sign = -1;
11             number = -number;
12         }
13 
14         StringBuffer strbuf = new StringBuffer(number + "");
15         strbuf.reverse();
16         String str = strbuf.toString();
17         long returnBuffer = sign * Long.parseLong(str);
18 
19         if(returnBuffer > Integer.MAX_VALUE || returnBuffer < Integer.MIN_VALUE)
20             return 0;
21         else
22             return (int)returnBuffer;
23     }
24 }