Valid Palindrome

2/20/2018 update
Better not include while within while. Use the original solution.
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11/30/2017 update
A more concise version is as follows.

 class Solution {  
   public boolean isPalindrome(String s) {    
     int i = 0, j = s.length() - 1;    
     while (i < j) {      
       while (i<j&&!Character.isLetterOrDigit(s.charAt(i))) i++;      
       while (i<j&&!Character.isLetterOrDigit(s.charAt(j))) j--;     
       if (Character.toLowerCase(s.charAt(i))!= Character.toLowerCase(s.charAt(j)))   
       {  
         return false;  
       }     
       i++;   
       j--;    
     }    
     return true; }   
 }  

Have to add i<j as condition when skip none letter or digit char, otherwise index would out of bound.

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思路：
这道题主要要熟悉java build-in 对操作string的一些函数。
String.toLowerCase().
Character.isLetterOrDigit().

实现：
先把s转换成lower case，这样不用在循环里面再每个转了。lintcode答案是刷题学生自己传上去的，很多不够优化。代码如下：

 public class Solution {  
   /*  
    * @param s: A string  
    * @return: Whether the string is a valid palindrome  
    */  
   public boolean isPalindrome(String s) {  
     int left=0;  
     int right=s.length()-1;  
     s=s.toLowerCase();  
     while(left<right)  
     {  
       char lc=s.charAt(left);  
       char rc=s.charAt(right);  
       if(!Character.isLetterOrDigit(lc))left++;  
       else if(!Character.isLetterOrDigit(rc))right--;  
       else if(lc!=rc)  
         return false;  
       else{  
       left++;  
       right--;  
       }  
     }  
     return true;  
   }  
 }