Word Ladder II leetcode java

题目：

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.

题解：

答案是http://www.1point3acres.com/bbs/thread-51646-1-1.html 上面
iostreamin写的。

我就直接贴过来就好，这道题多读读代码看明白。

代码：

  1    public ArrayList<ArrayList<String>> findLadders(String start, String end, HashSet<String> dict) {  
  2           
  3         HashMap<String, HashSet<String>> neighbours = new HashMap<String, HashSet<String>>();  
  4           
  5         dict.add(start);  
  6         dict.add(end);  
  7           
  8         // init adjacent graph          
  9         for(String str : dict){  
 10             calcNeighbours(neighbours, str, dict);  
 11         }  
 12           
 13         ArrayList<ArrayList<String>> result = new ArrayList<ArrayList<String>>();  
 14           
 15         // BFS search queue  
 16         LinkedList<Node> queue = new LinkedList<Node>();  
 17         queue.add(new Node(null, start, 1)); //the root has not parent and its level == 1 
 18           
 19         // BFS level  
 20         int previousLevel = 0;  
 21           
 22         // mark which nodes have been visited, to break infinite loop  
 23         HashMap<String, Integer> visited = new HashMap<String, Integer>();   
 24         while(!queue.isEmpty()){  
 25             Node n = queue.pollFirst();              
 26             if(end.equals(n.str)){   
 27                 // fine one path, check its length, if longer than previous path it's valid  
 28                 // otherwise all possible short path have been found, should stop  
 29                 if(previousLevel == 0 || n.level == previousLevel){  
 30                     previousLevel = n.level;  
 31                     findPath(n, result);                      
 32                 }else {  
 33                     // all path with length *previousLevel* have been found  
 34                     break;  
 35                 }                  
 36             }else {  
 37                 HashSet<String> set = neighbours.get(n.str);                   
 38                   
 39                 if(set == null || set.isEmpty()) continue;  
 40                 // note: I'm not using simple for(String s: set) here. This is to avoid hashset's  
 41                 // current modification exception.  
 42                 ArrayList<String> toRemove = new ArrayList<String>();  
 43                 for (String s : set) {  
 44                       
 45                     // if s has been visited before at a smaller level, there is already a shorter   
 46                     // path from start to s thus we should ignore s so as to break infinite loop; if   
 47                     // on the same level, we still need to put it into queue.  
 48                     if(visited.containsKey(s)){  
 49                         Integer occurLevel = visited.get(s);  
 50                         if(n.level+1 > occurLevel){  
 51                             neighbours.get(s).remove(n.str);  
 52                             toRemove.add(s);  
 53                             continue;  
 54                         }  
 55                     }  
 56                     visited.put(s,  n.level+1);  
 57                     queue.add(new Node(n, s, n.level + 1));  
 58                     if(neighbours.containsKey(s))  
 59                         neighbours.get(s).remove(n.str);  
 60                 }  
 61                 for(String s: toRemove){  
 62                     set.remove(s);  
 63                 }  
 64             }  
 65         }  
 66   
 67         return result;  
 68     }  
 69       
 70     public void findPath(Node n, ArrayList<ArrayList<String>> result){  
 71         ArrayList<String> path = new ArrayList<String>();  
 72         Node p = n;  
 73         while(p != null){  
 74             path.add(0, p.str);  
 75             p = p.parent;   
 76         }  
 77         result.add(path);  
 78     }  
 79   
 80     /* 
 81      * complexity: O(26*str.length*dict.size)=O(L*N) 
 82      */  
 83     void calcNeighbours(HashMap<String, HashSet<String>> neighbours, String str, HashSet<String> dict) {  
 84         int length = str.length();  
 85         char [] chars = str.toCharArray();  
 86         for (int i = 0; i < length; i++) {  
 87               
 88             char old = chars[i];   
 89             for (char c = 'a'; c <= 'z'; c++) {  
 90   
 91                 if (c == old)  continue;  
 92                 chars[i] = c;  
 93                 String newstr = new String(chars);                  
 94                   
 95                 if (dict.contains(newstr)) {  
 96                     HashSet<String> set = neighbours.get(str);  
 97                     if (set != null) {  
 98                         set.add(newstr);  
 99                     } else {  
                         HashSet<String> newset = new HashSet<String>();  
                         newset.add(newstr);  
                         neighbours.put(str, newset);  
                     }  
                 }                  
             }  
             chars[i] = old;  
         }  
     }  
       
     private class Node {  
         public Node parent;  //previous node
         public String str;  
         public int level;  
         public Node(Node p, String s, int l){  
             parent = p;  
             str = s;  
             level = l;  
         }  
     } 

 Reference：http://www.1point3acres.com/bbs/thread-51646-1-1.html