Valid Palindrome

验证回文字符串

最新推荐文章于 2024-01-04 12:53:14 发布

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本文介绍了一个简单的算法，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进行比较，有效处理了空字符串等特殊情况。

Valid Palindrome

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.

Answer:

思路十分简单，前后前后两个字符比较，直到正中间。

代码实现：

class Solution {
public:
	bool isPalindrome(string s) 
	{
		if (s.length() == 0)
			return true;
		int i = 0, j = s.length() - 1;
		while (i < j)
		{
			if (!isAlphanumeric(s.at(i)))
				i++;
			else if (!isAlphanumeric(s.at(j)))
				j--;
			else if (tolower(s.at(i)) ==tolower( s.at(j)))
			{
				i++; j--;
			}
			else
				return false;
		}
		return true;
	}

	bool isAlphanumeric(char a)
	{
		if (a >= 'a' && a <= 'z' || a >= 'A' && a <= 'Z' || a >= '0' && a <= '9')
			return true;
		else
			return false;
	}
};