15算法课程 125. Valid Palindrome

最新推荐文章于 2025-09-11 22:31:07 发布

QingJiuYou

最新推荐文章于 2025-09-11 22:31:07 发布

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本文介绍了一种方法来判断给定的字符串是否为回文，仅考虑字母和数字字符并忽略大小写。通过循环遍历字符串，过滤掉非字母数字字符，并利用大小写转换进行比较，最终确定字符串是否满足回文特性。

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.

solution:

循环遍历字符串，将非字母非数字的字符用''直接替换掉得到纯字母数字字符串，然后忽略大小写lower()或者upper()判断是否为回文

code:

class Solution {
public:
    bool isPalindrome(string s) {
        int left=0,right=s.size()-1;
        while(left<right)
        {
            while(left<right && isalnum(s[left]) == false)
                left++;
            while(left<right && isalnum(s[right]) == false )
                right--;
            if(left>=right) return true;
            if(toupper(s[left]) != toupper(s[right]))
                return false;
            left++,right--;
        }
        return true;
    }
};