Valid Palindrome leetcode

最新推荐文章于 2019-04-02 11:02:25 发布

weixin_33860147

最新推荐文章于 2019-04-02 11:02:25 发布

阅读量47

点赞数

CC 4.0 BY-SA版权

原文链接：http://www.cnblogs.com/sdlwlxf/p/5096994.html

本文介绍了一个C++函数，用于判断给定的字符串是否为回文，仅考虑字母数字字符并忽略大小写。通过迭代字符串的首尾来实现，跳过非字母数字字符，并将所有字符转换为小写进行比较。

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.

Subscribe to see which companies asked this question

bool isPalindrome(string s) {
    string::iterator iter1 = s.begin();
    string::iterator iter2 = s.end() - 1;
    while (iter1 <= iter2)
    {
        if (!isalpha(*iter1) && !isdigit(*iter1)) {
            iter1++;
            continue;
        }            
        if (!isalpha(*iter2) && !isdigit(*iter2)) {
            iter2--;
            continue;
        }
        char c1 = tolower(*iter1);
        char c2 = tolower(*iter2);
        if (c1 != c2)
            return false;
        iter1++;
        iter2--;
    }
    return true;
}

转载于:https://www.cnblogs.com/sdlwlxf/p/5096994.html