【leetcode】Palindrome Number

最新推荐文章于 2025-09-12 17:31:31 发布

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最新推荐文章于 2025-09-12 17:31:31 发布

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文章标签：数据结构与算法

原文链接：http://www.cnblogs.com/MrLJC/p/4241780.html

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。通过直接比较数字的最高位和最低位，避免了转换为字符串带来的空间开销。文章提供了两种实现方案并讨论了负数和溢出等特殊情况。

题目简述：

Determine whether an integer is a palindrome. Do this without extra space.
Some hints:
Could negative integers be palindromes? (ie, -1)

If you are thinking of converting the integer to string, note the restriction of using extra space.

You could also try reversing an integer. However, if you have solved the problem "Reverse Integer", you know that the reversed integer might overflow. How would you handle such case?

There is a more generic way of solving this problem.

解题思路：

首先最简单的就是想到直接转成字符串最前一个和最后一个比较，而且做出来效果还不错。不过题目说可以直接操作数字，那么我们也可以想办法每次取出数字的最高位和最低位进行比较，办法就是用一个base取最高位，每次取后base/100

class Solution:
    # @return a boolean
    def isPalindrome(self, x):
        if x < 0:
            return False
        else:
            s = str(x)
        l = len(s)
        for i in range(l/2):
            if s[i] != s[l-i-1]:
                return False
        return True

class Solution:
    # @return a boolean
    def isPalindrome(self, x):
        if x < 0:
            return False
        base = 1
        while x / base >= 10:
            base *= 10
        while  x > 0:
            left = x / base
            right = x % 10
            if left != right:
                return False
            x -= base * left
            x /= 10
            base /= 100
        return True

转载于:https://www.cnblogs.com/MrLJC/p/4241780.html