準(zhǔn)備開始刷題，然后希望把做過的題寫在這里，這樣既能加深印象，又能有機(jī)會(huì)找出自己潛在的問題。
題目如下：

Given a string, determine if it is a palindrome, considering only alphanumeric characters and ignoring cases.
Notice
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.

解題代碼如下：
思路很簡單，就是設(shè)置兩個(gè)指針，一個(gè)頭一個(gè)尾，一直對(duì)比到中間，如果遇到不一樣就return false。
當(dāng)然，題目還要求忽略空格和數(shù)字，所以這題需要用到Character對(duì)象的方法， 這個(gè)時(shí)間久遠(yuǎn)不用可能會(huì)忘，所以需要認(rèn)真看一下。

public class Solution {
/**
* @param s A string
* @return Whether the string is a valid palindrome
*/
public boolean isPalindrome(String s) {
          // Write your code here
          if( s.length() == 0){
          return true;
          }
          int left = 0;
          int right = s.length()-1;
          while(left < right){
                    if(!Character.isLetter(s.charAt(left)) && !Character.isDigit(s.charAt(left))){
                              left++;
                              continue;
                    }
                    if(!Character.isLetter(s.charAt(right)) && !Character.isDigit(s.charAt(right))){
                              right--;
                              continue;
                    }
                    if(Character.toUpperCase(s.charAt(left++)) != Character.toUpperCase(s.charAt(right--))){
                              return false;
                    }
          }
          return true;
          }
}