valid palindrome

最新推荐文章于 2026-01-05 22:36:07 发布

原创最新推荐文章于 2026-01-05 22:36:07 发布 · 1.4k 阅读

0 ·

CC 4.0 BY-SA版权

文章标签：

#C++ #vector #leetcode #string

LeetCode 专栏收录该内容

112 篇文章

订阅专栏

本文介绍了一种通过双指针法来判断字符串是否为回文的有效算法，该算法考虑了只包含字母数字字符的情况，并忽略了大小写差异。

题目：

Given a string, determine if it is a palindrome, considering only alphanumeric characters and ignoring cases.

For example,
"A man, a plan, a canal: Panama" is a palindrome.
"race a car" is not a palindrome.

Note:
Have you consider that the string might be empty? This is a good question to ask during an interview.

For the purpose of this problem, we define empty string as valid palindrome.

思路：

其实很简单，设置两个指针，头指针i 和尾指针j ，都向中间靠拢，判断有效地 s[i]和s[j]是否相等。不相等则返回false。

class Solution {
public:
    bool isPalindrome(string s) {
        // Start typing your C/C++ solution below
        // DO NOT write int main() function
        
        if(s.empty())
            return true;
        int i=0;
        int j=s.size()-1;
        while(i<=j)
        {
            if(!isalnum(s[i]))
            {
                i++;
                continue;
            }
            if(!isalnum(s[j]))
            {
                j--;
                continue;
            }
            if(s[i]==s[j])
            {
                i++;
                j--;
                continue;
            }
            else if(isalpha(s[i]) && isalpha(s[j]) && tolower(s[i])==tolower(s[j]))
                 {
                     i++;
                     j--;
                     continue;
                 }
                 else
                     return false;
        }
        return true;
    }
};

Run Status: Accepted!
Program Runtime: 52 milli secs