125. Valid Palindrome

最新推荐文章于 2024-01-04 12:53:14 发布

ShenYounger

最新推荐文章于 2024-01-04 12:53:14 发布

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本文链接：https://blog.youkuaiyun.com/wusecaiyun/article/details/47615251

本文介绍了一种算法，用于判断一个字符串是否为回文，只考虑字母数字字符并忽略大小写。通过双指针从两端向中间扫描的方式实现，跳过所有非字母数字字符。

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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.

两个指针首尾判断即可，跳过非字母数字字符，注意大小写，另外记住'A'要比'a'小32。代码如下：

class Solution {
public:
    bool isPalindrome(string s) {
        // IMPORTANT: Please reset any member data you declared, as
        // the same Solution instance will be reused for each test case.
        int len = s.length();
        if(len <= 1)return true;
        int i = 0, j = len - 1;
        while(i <= j)
        {
            if(isAlphanumeric(s[i]) == false)
                i++;
            else if(isAlphanumeric(s[j]) == false)
                j--;
            else 
            {
                if(isSame(s[i], s[j]) == true)
                    {i++; j--;}
                else return false;
            }
        }
        return true;
    }
    bool isAlphanumeric(char c)
    {
        if((c >= 'a' && c <= 'z')||(c >= 'A' && c <= 'Z')||(c >= '0' && c <= '9'))
            return true;
        else return false;
    }
    bool isSame(char a, char b)
    {
        if(a >= 'A' && a <= 'Z')
            a += 32;
        if(b >= 'A' && b <= 'Z')
            b += 32;
        return a == b;
    }
};