leetcode之Valid Palindrome-优快云博客

本文链接：https://blog.youkuaiyun.com/momentforver/article/details/26711089

本文介绍如何通过编程判断给定字符串是否为回文串，即正读反读都相同的字符串，忽略非字母数字字符并区分大小写。

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原题如下：

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.

这道题的难度不大，关键是要对当前字符是否是字母数字进行判断，这就要求能够掌握字母数字所对应的ASCII表，其中0-9所对应的ASCII为48-57，A-Z所对应的ASCII码为65-90，a-z所对应的ASCII码为97-122，为了简化判断，可以利用tranform（）方法将大写字母转化成小写字母，然后分别从前后两个方向比较即可。

class Solution {
public:
    bool isPalindrome(string s) {
		if(s.size() <= 1)
			return true;
		transform(s.begin(),s.end(),s.begin(),::tolower);		
		int left = 0,right = s.size() - 1;
		while(left < right){
			while(left < right && (s[left] < 48 || (s[left] > 57 && s[left] < 97) || s[left] > 122) )
				left++;
			while(right > left && (s[right] < 48 || (s[right] > 57 && s[right] < 97) || s[right] > 122))
				right--;
			if(s[left] == s[right])
			{
				left++;
				right--;
			}
			else				
				return false;			
		}		
		return true;
    }
};