[LeetCode]Valid Palindrome

最新推荐文章于 2019-11-04 14:27:48 发布

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最新推荐文章于 2019-11-04 14:27:48 发布

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原文链接：http://www.cnblogs.com/Rosanna/p/3463578.html

本文介绍了一个C++实现的算法，用于判断一个字符串是否为回文，仅考虑字母和数字字符并忽略大小写。通过双指针技巧从两端向中间遍历比较字符，有效处理了特殊字符过滤和大小写不敏感的问题。

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.

思考：注意判断语句是否正确。

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

转载于:https://www.cnblogs.com/Rosanna/p/3463578.html