125. Valid Palindrome

Character.isLetterOrDigit()

 

 1 class Solution {
 2     public boolean isPalindrome(String s) {
 3         if(s.length() == 0) return true;
 4         StringBuilder sb = new StringBuilder();
 5         for(int i = 0; i < s.length(); i++){
 6             if(Character.isLetterOrDigit(s.charAt(i))){
 7                 sb.append(s.charAt(i));
 8             }
 9         }
10         String str = sb.toString().toLowerCase();
11         sb = new StringBuilder(str);
12         StringBuilder sb1 = new StringBuilder(str);
13         sb.reverse();
14         
15         return sb.toString().equals(sb1.toString());
16         
17     }
18 }

 

posted @ 2018-10-26 06:04  jasoncool1  阅读(128)  评论(0)  编辑  收藏  举报