Leetcode练习－ Reverse Integer

最新推荐文章于 2025-08-08 00:00:33 发布

Flyinocean

最新推荐文章于 2025-08-08 00:00:33 发布

阅读量458

点赞数

CC 4.0 BY-SA版权

文章标签： leetcode python reverse integer

本文链接：https://blog.youkuaiyun.com/Flyinocean/article/details/38631679

本文介绍了一种将整数数字进行翻转的简单算法，并通过Python实现。文章讨论了正负号处理、溢出问题及特殊情况如末位为0的情况，并提供了一个具体的Python类和方法实现。

摘要生成于 C知道，由 DeepSeek-R1 满血版支持，前往体验 >

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).

分析一下：要考虑数字开始的正负，用一个flag保存＋／－。因为写python，变量integer就本身不强制类型，overflow的事情不用担心了，呵呵。异常抛出也是多余的在这里。思路就是每次把x对10取余，然后一步步把余数＊10，累积起来。没啥特别的。如果用c++ java 可能要考虑overflow，这估计是出题的最大目的？不确定。

class Solution:
    # @return an integer
    def reverse(self, x):
        res = 0
        flag = 1 if x >0 else -1
        x = abs(x)
        while x > 0:
            res = res * 10 + x % 10
            x /= 10 
        return sign * res