Leetcode: Reverse Integer

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最新推荐文章于 2025-08-08 00:00:33 发布

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分类专栏： Leetcode 文章标签： leetcode

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本文探讨了如何实现整数反转并处理边界情况，包括最后一位为0的整数和可能引发溢出的整数反转操作。文章还讨论了在不同场景下如何避免异常行为，确保函数的稳定性和正确性。

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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?

Throw an exception? Good, but what if throwing an exception is not an option? You would then have to re-design the function (ie, add an extra parameter).

除以10，存结果和余数，并计算，直到结果为0。一下代码没有考虑溢出的情况，如果溢出，得到的结果会和输入值符号相反，以此判断是否溢出，用另外一个参数来存最后一位值。

public class Solution {
    public int reverse(int x) {
        int res = 0;
        while (x != 0) {
            res = res * 10 + x % 10;
            x = x / 10;
        }
        
        return res;
    }
}