LeetCode-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.

Solution:

 1 public class Solution {
 2     public boolean isPalindrome(String s) {
 3         if (s.isEmpty()) return true;
 4 
 5         int p1 = 0, p2 = s.length()-1;
 6 　　　　 while (p1<s.length() && !( (s.charAt(p1)>='A' && s.charAt(p1)<='Z')  || (s.charAt(p1)>='a' && s.charAt(p1)<='z') || (s.charAt(p1)>='0' && s.charAt(p1)<='9') ) ) p1++;
 7　　　　　while (p2>=0 && !( (s.charAt(p2)>='A' && s.charAt(p2)<='Z')  || (s.charAt(p2)>='a' && s.charAt(p2)<='z') || (s.charAt(p2)>='0' && s.charAt(p2)<='9') ) ) p2--;
 8 
 9         while (p1<p2){
10             char c1 = s.charAt(p1);
11             char c2 = s.charAt(p2);
12             if (c1 == c2 || Math.abs((int) c1 - (int) c2) == Math.abs( (int) 'A' - (int) 'a' ) ){
13                 p1++;
14                 p2--;
15        　　　　 while (p1<s.length() && !( (s.charAt(p1)>='A' && s.charAt(p1)<='Z')  || (s.charAt(p1)>='a' && s.charAt(p1)<='z') || (s.charAt(p1)>='0' && s.charAt(p1)<='9') ) ) p1++;
16 　　　　　　　　 while (p2>=0 && !( (s.charAt(p2)>='A' && s.charAt(p2)<='Z')  || (s.charAt(p2)>='a' && s.charAt(p2)<='z') || (s.charAt(p2)>='0' && s.charAt(p2)<='9') ) ) p2--;
17 　　　　　　　} else return false;
18         }
19 
20         return true;
21     }
22 }

 

posted @ 2015-01-05 06:40  LiBlog  阅读(142)  评论(0)  编辑  收藏  举报