Problem Word Ladder II

Problem Description:

Given two words (start and end), and a dictionary, find all shortest transformation sequence(s) 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"]

Return

  [
    ["hit","hot","dot","dog","cog"],
    ["hit","hot","lot","log","cog"]
  ]

 

Note:

  • All words have the same length.
  • All words contain only lowercase alphabetic characters.

 

 

 Solution:
 1     public List<List<String>> findLadders(String start, String end, Set<String> dict) {
 2         List<List<String>> ret = new LinkedList<List<String>>();
 3         Map<String, List<String>> adjMap = new HashMap<String, List<String>>();
 4         
 5         List<Set<String>> layers = new LinkedList<Set<String>>();
 6         layers.add(0, new HashSet<String>());
 7         layers.add(1, new HashSet<String>());
 8         dict.add(start);
 9         dict.add(end);
10         int cur = 0;
11         int pre = 1;
12 
13         layers.get(cur).add(start);
14         // BFS 
15         while (! layers.get(cur).isEmpty() && ! layers.get(cur).contains(end)) {
16             cur = (cur == 0) ? 1 : 0;
17             pre = (pre == 0) ? 1 : 0;
18             for (String word : layers.get(pre)) dict.remove(word);
19 
20             layers.get(cur).clear();
21 for (String word: layers.get(pre)) {
22                 for (int i = 0; i < word.length(); ++i) {
23                     char[] charArray = word.toCharArray();
24                     for (char c = 'a'; c <= 'z'; ++c) {
25                         if (charArray[i] == c) continue;
26                         charArray[i] = c;
27                         String temp = new String(charArray);
28                         if (dict.contains(temp)) {
29                             if (! adjMap.containsKey(temp)) {
30                                 List<String> path = new LinkedList<String>();
31                                 adjMap.put(temp, path);
32                             }
33                             adjMap.get(temp).add(word);
34                             layers.get(cur).add(temp);
35                         }
36                     }
37                 }
38             }
39 
40         }
41 
42         if (layers.get(cur).isEmpty()) {
43             return ret;
44         } 
45 
46         List<String> path = new LinkedList<String> ();
47         buildPath(ret, adjMap, path, end, start);
48         return ret;
49     }
50   public void buildPath(List<List<String>> ret, Map<String, List<String>> adjMap, List<String> path, String cur, String start) {
51         if (start.equals(cur)) {
52             path.add(0, cur);
53             ret.add(path);
54             return;
55         }
56 
57         path.add(0, cur);
58         for (int i = 0; i < adjMap.get(cur).size(); ++i) {
59             List<String> copy = new LinkedList(path);
60             buildPath(ret, adjMap, copy, adjMap.get(cur).get(i), start);
61         }
62     }

 

posted @ 2014-06-29 14:52  HaruHaru  阅读(161)  评论(0)  编辑  收藏  举报