leetcode---valid-palindrome

最新推荐文章于 2018-09-15 16:55:00 发布

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最新推荐文章于 2018-09-15 16:55:00 发布

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本文介绍了一个C++实现的方法，用于检查给定字符串是否为回文，仅考虑字母数字字符并忽略大小写。通过实例演示了如何过滤掉非字母数字字符，并使用双指针技巧来比较字符串两端的字符。

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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.

class Solution {
public:
    string strupr(string &s)
    {
        for(int i=0; i<s.size(); i++)
            s[i] = toupper(s[i]);
        return s;
    }

    bool isPalindrome(string s) 
    {
        if(s.size() == 0)
            return true;
        s = strupr(s);
        int l = 0;
        int r = s.size() - 1;
        while(l <= r)
        {
            while(l<r && !isalpha(s[l]) && !isdigit(s[l]))
                ++l;
            while(l<r && !isalpha(s[r]) && !isdigit(s[r]))
                --r;
            if(s[l] != s[r])
                return false;
            ++l;
            --r;
        }
        return true;
    }
};