Shortest Palindrome

1. Title

Shortest Palindrome

2.   Http address

https://leetcode.com/problems/shortest-palindrome/

3. The question

Given a string S, you are allowed to convert it to a palindrome by adding characters in front of it. Find and return the shortest palindrome you can find by performing this transformation.

For example:

Given "aacecaaa", return "aaacecaaa".

Given "abcd", return "dcbabcd".

4. My code (AC)

 1  //Accepted
 2       public  String shortestPalindrome0(String s) {
 3           if( s == null || s.length() <= 1 || this.isPalindrom(s))
 4           {
 5               return s;
 6           }
 7           int len = s.length();
 8           int mid = len /2;
 9           for(int i = mid; i >= 0; i--)
10           {
11               StringBuilder sb = new StringBuilder("");
12               if(checkIsPalindrom(s,i,sb))
13               {
14                  // int dis = (len-i) - (i);
15                   //System.out.println("outer" + sb.toString());
16                   int index = Integer.parseInt(sb.toString());
17                   
18                   StringBuilder strB = new StringBuilder(s.substring(index));
19                   return strB.reverse().toString() + s;
20               }
21           }
22           return "error";
23       }
24       // 
25     private boolean checkIsPalindrom(String s, int mid, StringBuilder sb) {
26         // TODO Auto-generated method stub
27         int len = s.length();
28         int i = mid - 1;
29         int j = mid;
30         if( mid < 1){
31             sb.append(1);
32             return true;
33         }
34         
35         
36         while( j < len && i >= 0)
37         {
38             if(s.charAt(j) != s.charAt(i))
39             {
40                 break;
41             }
42             j++;
43             i--;
44         }
45         
46         if( j < len && i >= 0)
47         {
48             i = mid - 1;
49             j = mid + 1;
50             if( j >= len)
51             {
52                 return false;
53             }
54             while( j < len && i >= 0)
55             {
56                 if(s.charAt(j) != s.charAt(i))
57                 {
58                     return false;
59                 }
60                 j++;
61                 i--;
62             }
63         }
64         
65         sb.append(j);
66         return true;
67     }
68 
69     public   boolean isPalindrom(String s){
70         if( s == null || s.length() <= 1)
71             return true;
72         int len = s.length();
73         int i = 0;
74         int j = len -1;
75         while( i < j)
76         {
77             if( s.charAt(i) != s.charAt(j)){
78                 return false;
79             }
80             i++;
81             j--;
82         }
83         return true;
84     }

 

posted @ 2015-11-01 18:19  ordi  阅读(132)  评论(0)  编辑  收藏  举报