Valid Palindrome

最新推荐文章于 2024-01-04 12:53:14 发布

原创最新推荐文章于 2024-01-04 12:53:14 发布 · 188 阅读

0 ·

CC 4.0 BY-SA版权

LeetCode之旅同时被 2 个专栏收录

65 篇文章

订阅专栏

c++

29 篇文章

订阅专栏

本文介绍了一种判断字符串是否为回文的有效算法，该算法仅考虑字母数字字符并忽略大小写差异。通过双指针技术从两端向中间扫描，跳过非字母数字字符，实现了高效判断。

摘要生成于 C知道，由 DeepSeek-R1 满血版支持，前往体验 >

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.

class Solution {
public:
    bool isPalindrome(string s) {
        if(s.length()<=1)
            return true;
            
        int start = 0;
        int end = s.length()-1;
        while(start<end){
            while(start<end&&!isValid(s[start]))
                start++;
            while(start<end&&!isValid(s[end]))
                end--;
            if(start>=end)
                break;
            if(s[start] != s[end] && abs(s[start] - s[end]) != abs('A'-'a')){
                return false;
            }else{
                start++;
                end--;
            }
        }
        return true;
    }
    
    bool isValid(char c){
        if(c>='a'&&c<='z')
            return true;
        if(c>='A'&&c<='Z')
            return true;
        if(c>='0'&&c<='9')
            return true;
        return false;
    }
};