[LeetCode126] Valid Palindrome

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.

Analysis:

Use Two pointers, scan from end to start.

Java

public boolean isPalindrome(String s) {
        s = s.trim();
		if(s.length()<=0) return true;
		int start = 0;
		int end = s.length()-1;
		while(start<end){
			while(start<end && !(Character.isLetter(s.charAt(start))||Character.isDigit(s.charAt(start)))) start++;
			while(start<end &&  !(Character.isLetter(s.charAt(end))||Character.isDigit(s.charAt(end)))) end--;
			if(s.charAt(start)==s.charAt(end) || s.charAt(start)-'a'==s.charAt(end)-'A'||s.charAt(start)-'A' == s.charAt(end)-'a' ){
			start++;
			end--;
			}else {
				break;
			}
		}
		if(start>=end)
			return true;
		else {
			return false;
		}
    }

Another one:

public boolean isPalindrome(String s) {
        if(s.isEmpty()) return true;
        int len = s.length();
        int start = 0;
        int end = len-1;
        while(start<=end){
        	if(!isAlp(s.charAt(start))){
        		start++;
        		continue;
        	}
        	if(!isAlp(s.charAt(end))){
        		end--;
        		continue;
        	}
        	if(s.charAt(start)==s.charAt(end) || Math.abs(s.charAt(start)-s.charAt(end))==32){
        		start++;
				end--;
        	}else
        		return false;
        	
        }
        return true;
    }
	public boolean isAlp(char c){
		if((c<='z'&&c>='a')||(c<='Z'&&c>='A')|| (c>='0'&&c<='9'))
			return true;
		else
			return false;
	}

c++

bool isPalindrome(string s) {
      if(s.empty()) return true;
    int len = s.length();
    int r = len-1;
    int l = 0;
    while(l<=r){
        if((s[l] >= 'a' && s[l]<='z') || (s[l]>='A' && s[l]<='Z')||(s[l])>='0'&& s[l]<='9'){
            if((s[r] >= 'a' && s[r]<='z') || (s[r]>='A' && s[r]<='Z')||(s[r]>='0' && s[r]<='9')){
                if(s[l] == s[r] || s[l]-'a'==s[r]-'A' || s[l]-'A' == s[r]-'a'){
                    l++;
                    r--;
                }
                else
                    return false;
            }else{
                r--;
                continue;
            }
        }

        else{
            l++;
            continue;
        }
    }
    return true;
    }