leetcode -- Word Ladder

Given two words (start and end), and a dictionary, find the length of shortest transformation sequence from start to end, such that:

1. Only one letter can be changed at a time
2. Each intermediate word must exist in the dictionary

For example,

Given:
start = "hit"
end = "cog"
dict = ["hot","dot","dog","lot","log"]

As one shortest transformation is "hit" -> "hot" -> "dot" -> "dog" -> "cog",
return its length 5.

Note:

• Return 0 if there is no such transformation sequence.
• All words have the same length.
• All words contain only lowercase alphabetic characters.

 

bug code

 1 public class Solution {
 2     public int ladderLength(String start, String end, HashSet<String> dict) {
 3         // Start typing your Java solution below
 4         // DO NOT write main() function
 5         if(dict.size() == 0 || start.equals(end)){
 6             return 0;
 7         }
 8         //dict.add(start);
 9         dict.remove(start);
10         dict.add(end);
11         int result = getLen(start, end, dict, 0);
12         if(result == 0){
13             return result;
14         } 
15         return result + 1; 
16     }
17     
18     public int getLen(String start, String end, HashSet<String> dict, int len){
19         if(start.equals(end)){
20             return len;
21         }
22         
23         HashSet<String> neighbors = getNeighbors(start, dict);
24         int maxLen = len;
25         for(String n : neighbors){
26             dict.remove(n);
27             int tmp = getLen(n, end, dict, len + 1);
28             if(tmp > maxLen){
29                 maxLen = tmp;
30             }
31             dict.add(n);
32         }
33         return maxLen;
34     }
35     
36     public HashSet<String> getNeighbors(String start, HashSet<String> dict){
37         HashSet<String> result = new HashSet<String>();
38         for(int i = 0; i < start.length(); i++){
39             StringBuffer sb = new StringBuffer();
40             if(i == 1){
41                 sb.append(start.substring(0, 1));
42             } else if(i == 0){
43                 
44             } else {
45                 sb.append(start.substring(0, i));
46             }
47             for(char c = 'a'; c <= 'z'; c++){
48                 if(c != start.charAt(i)){
49                     sb.append(c).append(start.substring(i + 1));
50                     if(dict.contains(sb.toString())){
51                         result.add(sb.toString());
52                     }
53                 }
54             }
55         }
56         return result;
57     }
58 }

View Code