**Word Ladder II

Given two words (beginWord and endWord), and a dictionary's word list, find all shortest transformation sequence(s) from beginWord to endWord, such that:

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

For example,

Given:
beginWord = "hit"
endWord = "cog"
wordList = ["hot","dot","dog","lot","log"]

Return

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

The solution contains two steps 1 Use BFS to construct a graph. 2. Use DFS to construct the paths from end to start.Both solutions got AC within 1s.

The first step BFS is quite important. I summarized three tricks

1) Using a MAP to store the min ladder of each word, or use a SET to store the words visited in current ladder, when the current ladder was completed, delete the visited words from unvisited. That's why I have two similar solutions.

2) Use Character iteration to find all possible paths. Do not compare one word to all the other words and check if they only differ by one character.

3) One word is allowed to be inserted into the queue only ONCE. See my comments.

public class Solution {
    Map<String,List<String>> map;
    List<List<String>> results;
    public List<List<String>> findLadders(String start, String end, Set<String> dict) {     
        results= new ArrayList<List<String>>();
        if (dict.size() == 0)
            return results;

        int min=Integer.MAX_VALUE;

        Queue<String> queue= new LinkedList<String>();
        queue.add(start);

        map = new HashMap<String,List<String>>();

        Map<String,Integer> ladder = new HashMap<String,Integer>();
        for (String string:dict)
            ladder.put(string, Integer.MAX_VALUE);
        ladder.put(start, 0);

        dict.add(end);
        //BFS: Dijisktra search
        while (!queue.isEmpty()) {

            String word = queue.poll();

            int step = ladder.get(word)+1;//'step' indicates how many steps are needed to travel to one word. 

            if (step>min) break;

            for (int i = 0; i < word.length(); i++){
               StringBuilder builder = new StringBuilder(word); 
                for (char ch='a';  ch <= 'z'; ch++){
                    builder.setCharAt(i,ch);
                    String new_word=builder.toString();             
                    if (ladder.containsKey(new_word)) {

                        if (step>ladder.get(new_word))//Check if it is the shortest path to one word.
                            continue;
                        else if (step<ladder.get(new_word)){
                            queue.add(new_word);
                            ladder.put(new_word, step);
                        }else;// It is a KEY line. If one word already appeared in one ladder,
                              // Do not insert the same word inside the queue twice. Otherwise it gets TLE.

                        if (map.containsKey(new_word)) //Build adjacent Graph
                            map.get(new_word).add(word);
                        else{
                            List<String> list= new LinkedList<String>();
                            list.add(word);
                            map.put(new_word,list);
                            //It is possible to write three lines in one:
                            //map.put(new_word,new LinkedList<String>(Arrays.asList(new String[]{word})));
                            //Which one is better?
                        }

                        if (new_word.equals(end))
                            min=step;

                    }//End if dict contains new_word
                }//End:Iteration from 'a' to 'z'
            }//End:Iteration from the first to the last
        }//End While

        //BackTracking
        LinkedList<String> result = new LinkedList<String>();
        backTrace(end,start,result);

        return results;        
    }
    private void backTrace(String word,String start,List<String> list){
        if (word.equals(start)){
            list.add(0,start);
            results.add(new ArrayList<String>(list));
            list.remove(0);
            return;
        }
        list.add(0,word);
        if (map.get(word)!=null)
            for (String s:map.get(word))
                backTrace(s,start,list);
        list.remove(0);
    }
}

Note:

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

reference：

https://leetcode.com/discuss/9523/share-two-similar-java-solution-that-accpted-by-oj

https://chazyhabit.wordpress.com/2014/07/25/word-ladder-ii-leetcode-18/