Given two words (start and end), and a dictionary, find all shortest transformation sequence(s) from start to end, such that:
- Only one letter can be changed at a time
- 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 }