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.

验证一个给定的字符串是否为回文，用两个指针分别指向字符的首尾，判断是否相同，相同就的都向中间移动1位，判断下一组，左指针小于右指针就一直循环。遇到标点符号就跳过，处理下一个。遇到大写字母就转换成小写字母。

Java：

class Solution {

    public boolean isPalindrome(String s) {

        char[] chs = s.toCharArray();

        int left = 0, right = s.length() - 1;

        while (left <= right) {

            while (left < right && !Character.isLetterOrDigit(chs[left]))

                left++;

            while (left < right && !Character.isLetterOrDigit(chs[right]))

                right--;

            if (Character.toLowerCase(chs[left++]) != Character.toLowerCase(chs[right--]))

                return false;

        }

        return true;

    }

}

Java:

public class Solution {

    public boolean isPalindrome(String s) {

        int size = s.length(), i = 0, j = size - 1;

        s = s.toLowerCase();

        while (i < j) {

            if (!(s.charAt(i) >= 'a' && s.charAt(i) <= 'z') && !(s.charAt(i) >= '0' && s.charAt(i) <= '9')) {

                i++;

            }

            else if (!(s.charAt(j) >= 'a' && s.charAt(j) <= 'z') && !(s.charAt(j) >= '0' && s.charAt(j) <= '9')) {

                j--;

            }

            else {

                if (s.charAt(i) != s.charAt(j)) return false;

                i++;

                j--;

            }

        }

        return true;

    }

}

Python：

class Solution:

    def isPalindrome(self, s):

        i, j = 0, len(s) - 1

        while i < j:

            while i < j and not s[i].isalnum():

                i += 1

            while i < j and not s[j].isalnum():

                j -= 1

            if s[i].lower() != s[j].lower():

                return False

            i, j = i + 1, j - 1

        return True

C++：

class Solution {

public:

    bool isPalindrome(string s) {

        int left = 0, right = s.size() - 1 ;

        while (left < right) {

            if (!isalnum(s[left])) ++left;

            else if (!isalnum(s[right])) --right;

            else if ((s[left] + 32 - 'a') %32 != (s[right] + 32 - 'a') % 32) return false;

            else {

                ++left; --right;

            }

        }

        return true;

    }

};

C++：

class Solution {

public:

    bool isPalindrome(string s) {

        int l = 0, r = s.size() - 1;

        while(l <= r){

            while(!isalnum(s[l]) && l < r) l++;

            while(!isalnum(s[r]) && l < r) r--;

            if(toupper(s[l]) != toupper(s[r])) return false;

            l++, r--;

        }

        return true;

    }

};

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