Given a string, determine if it is a palindrome, considering only alphanumeric characters and ignoring cases.只考虑字母数字，并忽略大小写

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.

Example
"A man, a plan, a canal: Panama" is a palindrome.

"race a car" is not a palindrome.

Challenge
O(n) time without extra memory.

keyword: 双指针
方法一：
100% test cases passedTotal runtime 180 ms
Your submission beats 96.20% Submissions!

public class Solution {
    public boolean isPalindrome(String s) {
        int len=s.length();
        char lft;
        char rht;
        if(len==0)
            return true;
        int lo=0;
        int hi=len-1;
        while(lo<hi){
            lft=s.charAt(lo);
            rht=s.charAt(hi);
            if(lft<'0' || (lft>'9' && lft<'A') || (lft>'Z' && lft<'a') || lft>'z'){
                lo++;
                continue;
            }
            if(rht<'0' || (rht>'9' && rht<'A') || (rht>'Z' && rht<'a') || rht>'z'){
                hi--;
                continue;
            }
            if(lft==rht || lft==rht+'a'-'A' || lft==rht-'a'+'A'){
                lo++;
                hi--;
            }else{
                return false;
            }
        }
        return true;
    }
}

方法二：利用Java库函数
toLowerCase(), Character.isLetterOrDigit()

100% test cases passedTotal runtime 184 ms
Your submission beats 66.60% Submissions!

public class Solution {
    public boolean isPalindrome(String s) {
        int len=s.length();
        char lft;
        char rht;
        if(len==0)
            return true;
        int lo=0;
        int hi=len-1;
        s=s.toLowerCase();
        while(lo<hi){
            lft=s.charAt(lo);
            rht=s.charAt(hi);
            if(!Character.isLetterOrDigit(lft)){
                lo++;
                continue;
            }
            if(!Character.isLetterOrDigit(rht)){
                hi--;
                continue;
            }
            if(lft==rht){
                lo++;
                hi--;
            }else{
                return false;
            }
        }
        return true;
    }
}