LeetCode第七题（Reverse Integer）

最新推荐文章于 2024-11-19 23:58:49 发布

泳冠三军

最新推荐文章于 2024-11-19 23:58:49 发布

阅读量219

点赞数

CC 4.0 BY-SA版权

分类专栏： LeetCode

本文链接：https://blog.youkuaiyun.com/ax478/article/details/79874912

LeetCode 专栏收录该内容

9 篇文章

订阅专栏

本文介绍了一个Python实现的算法，该算法可以将32位带符号整数的数字进行翻转。提供了详细的示例输入输出，并考虑了翻转过程中可能发生的溢出情况。

摘要生成于 C知道，由 DeepSeek-R1 满血版支持，前往体验 >

Reverse Integer

Given a 32-bit signed integer, reverse digits of an integer.

Example 1:

Input: 123
Output:  321

Example 2:

Input: -123
Output: -321

Example 3:

Input: 120
Output: 21

Note: Assume we are dealing with an environment which could only hold integers within the 32-bit signed integer range. For the purpose of this problem, assume that your function returns 0 when the reversed integer overflows.

Python代码：

class Solution:
    def overflow(self, x):
        """
        :type x: int
        :rtype: bool
        """
        if x > 0xFFFFFFFF or -x > 0xFFFFFFFF:
            return True
        return False

    def reverse(self, x):
        """
        :type x: int
        :rtype: int
        """
        MAX = 0x7FFFFFFF
        MIN = 0x80000000
        mask = 0xFFFFFFFF
        while (self.overflow(x) and x >= 0):
            x -= (mask + 1)
        while (self.overflow(x) and x < 0):
            x += (mask + 1)
        if x >= MIN and x <= mask:    #x为负整数
            x = -((x ^ mask) + 1)       #对x取二进制补码
        if x >= 0 and x <= MAX:         #x为正整数
            x = int(str(x)[::-1])
            if x > MAX:             #若反转后大于MAX，则溢出
                x = 0
        elif x < 0 and -x <= MIN:
            x = -int(str(x)[:0:-1])
        else:
            x = 0
        if (self.overflow(x) or (x > 0 and x > MAX) or (x < 0 and -x > MIN)):
            x = 0
        return x

运行结果