4.Valid Palindrome

Description:
Given a string, determine if it is a palindrome, considering only alphanumeric characters and ignoring cases.

Note:
For the purpose of this problem, we define empty string as valid palindrome

Leetcode link

solution 1
思路很简单，利用两个下标，一个从前一个从后，依次往前比较，但是超时了。

class Solution {
public:
   if ((s[j] >= 65 && s[j] <= 90) || (s[j] >= 97 && s[j] <= 122))
   {
	   if (s[i] == s[j] || s[i] - s[j] == 32 || s[j] - s[i] == 32)
	   {
			i++; j--;
			continue;
	   }
	   else
			return 0;
   }
   else if (s[j] == ' ' || !isalnum(s[j]))
	   j--;
	}
	else if (s[i] == ' '||!isalnum(s[i]))
	   i++;
	else
	   return 0;
  }
  return 1;
  }
};

solution 2
思路不变，用一个tolower来转换大小写，isalnum()判断是否为字母

class Solution {
public:
   bool isPalindrome(string s) {
	int j = s.length() - 1;
	if (j == -1 || j == 0)
		return 1;
	else
	{
		int i = 0;
		while (i < j)
		{
			char a=tolower(s[i]);
			char b=tolower(s[j]);
            if(a==b)
            {
                i++;j--;
                continue;
            }
            else if(a==' '||!isalnum(a))
                i++;
            else if(b==' '||!isalnum(b))
                j--;
            else
                return 0;
		}
		return 1;
	}
   }
};

做出来之后有一种蜜汁郁闷。
被整自闭了。
用例“lbl”,我记住了！