LeetCode125——Valid Palindrome_leet code oj 125. valid palindrome java-优快云博客

本文链接：https://blog.youkuaiyun.com/booirror/article/details/43197845

本文提供了一种方法来判断一个字符串是否为回文，仅考虑字母数字字符，忽略大小写和空格。同时，讨论了空字符串作为有效回文的情况，并在代码实现中详细阐述了如何过滤非字母数字字符。

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.

题目大意

给定一个字符串，判断是不是回文，只考虑字母数字字符，并不顾大小写。另外，空串被认为是有效的回文。

难度系数：容易

实现

bool isAlphanumeric(const char c) {
    if ('A' <= c && c <= 'Z')
        return true;
    else if ('a' <= c && c <= 'z')
        return true;
    else if ('0' <= c && c <= '9')
        return true;
    return false;
}

char toLower(const char c)
{
    if ('A' <= c && c <= 'Z')
        return 'a' + c - 'A';
    return c;
}

bool isPalindrome(string s)
{
    if (s.empty())
        return true;
    int begin = 0;
    int end = s.size() - 1;
    while (true) {
        while (begin < s.size() && !isAlphanumeric(s[begin])){
            begin++;
        }
        while (end >= 0 && !isAlphanumeric(s[end])) {
            end--;
        }
        if (end <= begin)
            return true;
        if (toLower(s[begin]) != toLower(s[end])){
            return false;
        }
        begin++;
        end--;
    }
}