Valid Palindromed

iven 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.

拿这个练习了下stack。注意top和pop。pop是不返回元素的。

class Solution {
public:
    bool isPalindrome(string s) {
        int l = s.length();
        stack<char> st;
        for(int i = 0 ; i < l ; i++)
        {
            if(s[i] >= 'a' && s[i] <= 'z'|| s[i] >='0'&&s[i]<='9')st.push(s[i]);
            else if(s[i] >= 'A' && s[i] <= 'Z') st.push(s[i]-'A'+'a');
            
        }
        
        for(int i = 0 ; i < l ; i++)
        {
            if(s[i] >= 'a' && s[i] <= 'z' || s[i] >='0'&&s[i]<='9')
            {
                char temp = st.top();
                st.pop();
                if(temp != s[i]) return 0;
            }
            else if(s[i] >= 'A' && s[i] <= 'Z')
            {
                char temp = st.top() -'a'+'A';
                st.pop();
                if(temp != s[i]) return 0;
            }
        }
        return 1;
    }
};

　　这个题有更优的解法，也不用栈。

     左右两个指针，分别指到第一个合法字符之后开始判断，之后一个向左一个向右，如此反复指到两个指针相遇。