[LeetCode]Valid Palindrome

很简单的判断回文的题目，双指针，跳过不考虑的字符即可，代码如下：

	class Solution {
	public:
	bool isPalindrome(string s) {
	int len = s.size(), i = 0, j = len - 1;
	while(true)
	{
	while(i < len && !isalnum(s[i]))++i;
	while(j >= 0 && !isalnum(s[j]))--j;
	if(i >= j)break;
	if(tolower(s[i]) != tolower(s[j]))return false;
	++i;
	--j;
	}
	return true;
	}
	};

view raw Valid Palindrome.cpp hosted with ❤ by GitHub