[LeetCode]125. 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.

Accepted pretty Java solution(271ms)

public class Solution {    public boolean isPalindrome(String s) {        if (s.isEmpty()) {        return true;        }        int head = 0, tail = s.length() - 1;        char cHead, cTail;        while(head <= tail) {        cHead = s.charAt(head);        cTail = s.charAt(tail);        if (!Character.isLetterOrDigit(cHead)) {        head++;        } else if(!Character.isLetterOrDigit(cTail)) {        tail--;        } else {        if (Character.toLowerCase(cHead) != Character.toLowerCase(cTail)) {        return false;        }        head++;        tail--;        }        }                return true;    }}

My three line java solution

public class Solution {    public boolean isPalindrome(String s) {        String actual = s.replaceAll("[^A-Za-z0-9]", "").toLowerCase();        String rev = new StringBuffer(actual).reverse().toString();        return actual.equals(rev);    }}

Passed clean c++ code

bool isPalindrome(string s) {int start=0, end=s.length()-1;while(start<end) {if (!isalnum(s[start])) start++;else if (!isalnum(s[end])) end--;else {if (tolower(s[start++])!=tolower(s[end--])) return false;}}return true;}

C版本解法：

bool isPalindrome(char* s) {
 int start = 0, end = strlen(s) - 1;
 while(start < end) {
     if(!isalnum(s[start]))
        start++;
     else if(!isalnum(s[end]))
        end--;
     else
        if(tolower(s[start++]) != tolower(s[end--]))
            return false;
 }
 return true;
}