[leetcode]126. Word Ladder II单词接龙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 transformed word must exist in the word list. Note that beginWord is not a transformed word.

Note:

• Return an empty list if there is no such transformation sequence.
• All words have the same length.
• All words contain only lowercase alphabetic characters.
• You may assume no duplicates in the word list.
• You may assume beginWord and endWord are non-empty and are not the same.

Example 1:

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

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

Example 2:

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

Output: []

Explanation: The endWord "cog" is not in wordList, therefore no possible transformation.

代码

class Solution {
    public List<List<String>> findLadders(String beginWord, String endWord, List<String> wordList) {
        // bfs
        HashSet<String> dict = new HashSet<>(wordList);
        dict.add(beginWord);
        HashMap<String, Integer> distance = new HashMap<>();
        HashMap<String, List<String>> adj = new HashMap<>();
        bfs(beginWord, endWord, dict, adj, distance);   
        //dfs
        List<List<String>> result = new ArrayList<>();
        List<String> path = new ArrayList<>();
        path.add(beginWord);
        dfs(beginWord, endWord, result, path, dict, adj, distance);
        return result;
    }
    
    public List<String> getNeighbor(String s, HashSet<String> dict) {
        List<String> list = new ArrayList<>(); 
        for (int i = 0; i < s.length(); i++) {
            char[] charArray = s.toCharArray();
            for (char c = 'a'; c <= 'z'; c++) { 
                if (charArray[i] == c)  continue;
                charArray[i] = c;
                String temp = new String(charArray);
                if (dict.contains(temp)) {
                    list.add(temp);
                }       
            }
        }
        return list;
    }
    
    public void bfs(String beginWord, 
                    String endWord, 
                    HashSet<String> dict, 
                    HashMap<String, List<String>> adj, 
                    HashMap<String, Integer> distance) {
        
        for (String word : dict) {
            adj.put(word, new ArrayList<String>());
        }  
        Queue<String> queue = new LinkedList<>();
        queue.add(beginWord);
        distance.put(beginWord, 0);
        while (!queue.isEmpty()) {
            String curr = queue.remove();
            int level = distance.get(curr);
            List<String> neighbor = getNeighbor(curr, dict);
            for (String nei : neighbor) {
                adj.get(curr).add(nei);
                if (!distance.containsKey(nei)) {
                    distance.put(nei, level + 1);
                    if (!endWord.equals(nei)) {
                        queue.add(nei);
                    }
                }
            }
        }
    }
    
    public void dfs(String curr,
                    String end, 
                    List<List<String>> result, 
                    List<String> path, 
                    HashSet<String> dict, 
                    HashMap<String, List<String>> adj, 
                    HashMap<String, Integer> distance) {
        
        if (curr.equals(end)) {
            result.add(new ArrayList<>(path));
            return;
        }
        
        for (String nei : adj.get(curr)) {
            path.add(nei);
            if (distance.containsKey(nei) && distance.get(nei) == distance.get(curr) + 1) {
                dfs(nei, end, result, path, dict, adj, distance);
            }
            path.remove(path.size() - 1);
        }
    }   
}