125. Valid Palindrome

最新推荐文章于 2024-01-04 12:53:14 发布

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最新推荐文章于 2024-01-04 12:53:14 发布

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本文介绍了一种通过两端向中间扫描的方式判断字符串是否为回文的方法，并考虑了大小写及非字母数字字符的问题。

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1.Question

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.

2.Code

class Solution {
public:
    bool isPalindrome(string s) {
        if(s.empty()) return true;
        int j = s.size() - 1;
        int i = 0;
        while(i < j)
        {
            char front = s[i], tail = s[j];
            while((front >'z' || front < 'a') && (front > '9' || front < '0') && i < j)
            {
                if(front <= 'Z' && front >= 'A') front = front - 'A' + 'a';
                else front = s[++i];
            }
            while((tail >'z' || tail < 'a') && (tail > '9' || tail < '0') && i < j)
            {
                if(tail <= 'Z' && tail >= 'A') tail = tail - 'A' + 'a';
                else tail = s[--j];
            }
            if(i < j)
            {
                if(front != tail) return false;
                else i++, j--;
            }
        }
        return true;
    }
};

3.Note
a. 从两端往中间找字母数字，然后进行比较是否相等。终止条件是i >= j。

b. 如果找到的是大写字母则转换为小写字母。