Word Ladder II

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.

和word ladder I一样，采用BFS + recurise来做。

 1 public class Solution {

 2     int len;

 3     int count;

 4     boolean sig;

 5     ArrayList<ArrayList<String>> result;

 6     HashSet<String> used;

 7     LinkedList<String> queue;

 8     String end;

 9     HashSet<String> dict;

10     public ArrayList<ArrayList<String>> findLadders(String start, String end, HashSet<String> dict) {

11         // Start typing your Java solution below

12         // DO NOT write main() function

13         len = start.length();

14         count = 0;

15         sig = false;

16         result = new ArrayList<ArrayList<String>>();

17         used = new HashSet<String>();

18         this.end = end;

19         this.dict = dict;

20         queue = new LinkedList<String>();

21         queue.add(start);

22         queue.add("-1");

23         if(len != end.length() || len == 0) return result;

24         ArrayList<String> row = new ArrayList<String>();

25         row.add(start);

26         getRows(row);

27         return result;

28     }

29     public void getRows(ArrayList<String> row){

30         if(queue.size() != 1){

31             String cur = queue.poll();

32             if(cur.equals("-1")){

33                 if(sig) return;

34                 count ++;

35                 queue.add("-1");

36             }else{

37                 used.add(cur);

38                 for(int i = 0; i < len; i ++)

39                     for(char c = 'a'; c < 'z' + 1; c ++){

40                         if(c == cur.charAt(i)) continue;

41                         char[] ts = cur.toCharArray();

42                         ts[i] = c;

43                         String test = String.valueOf(ts);

44                         if(dict.contains(test) && !used.contains(test)){

45                             used.add(test);

46                             queue.add(test);

47                         }

48                         if(test.equals(end)){ 

49                             row.add(end); 

50                             result.add(row);

51                             return;

52                         }

53                         ArrayList<String> rowa = new ArrayList<String>(row);

54                         rowa.add(test);

55                         getRows(rowa);

56                 }

57             }

58         }

59     }

60 }