9. Palindrome Number Leetcode Python 2016 new Season

hyperbolechi

于 2016-01-04 10:33:17 发布

阅读量287

点赞数

CC 4.0 BY-SA版权

分类专栏： leetcode 2016刷题

本文链接：https://blog.youkuaiyun.com/hyperbolechi/article/details/50454467

leetcode 同时被 2 个专栏收录

166 篇文章

订阅专栏

2016刷题

45 篇文章

订阅专栏

本文介绍了一种不使用额外空间判断整数是否为回文数的方法。通过将整数反转并与原数比较来实现，时间复杂度为O(log(n))。

摘要生成于 C知道，由 DeepSeek-R1 满血版支持，前往体验 >

Determine whether an integer is a palindrome. Do this without extra space.

click to show spoilers.

Show Tags

Show Similar Problems

The problem requirement doesn't make sense, we still need constant space to do this problem. the time complexity is O(log(n)).

class Solution(object):
    def isPalindrome(self, x):
        """
        :type x: int
        :rtype: bool
        """
        tmp_x = x
        residual = 0
        while x >= 1:
            rest = x % 10
            residual = residual * 10 + rest
            x /= 10
        if tmp_x == residual:
            return True
        return False