Valid Palindrome leetcode

最新推荐文章于 2019-04-02 11:02:25 发布

转载最新推荐文章于 2019-04-02 11:02:25 发布 · 52 阅读

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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。

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.

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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;
}