Leetcode 126. 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 transformed word must exist in the word list. Note that beginWord is not a transformed word.

For example,

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

Return

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

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.

用 一个哈希表 path 来 储存 word 和 word 所在层。用bfs 所有符合条件的 单词， 添加到 path 里。用dfs 从 end 往回找， 因为 endword 在 wordlist 里， beginword 不在wordlist 里

HashMap<String,Integer> path = new HashMap<String,Integer>();    public List<List<String>> findLadders(String beginWord, String endWord, List<String> wordList) {        List<List<String>> res = new ArrayList<>();        List<String> tmp = new ArrayList<>();        Set<String> set = new HashSet<>(wordList);        if (beginWord == null || endWord == null || !set.contains(endWord)) return res;        bfs(beginWord, endWord, set);        dfs(endWord, beginWord, set, tmp, res);        return res;    }    private void dfs(String beginWord, String endWord, Set<String> set, List<String> tmp_path, List<List<String>> res) {        if (beginWord.equals(endWord)) {            tmp_path.add(beginWord);            Collections.reverse(tmp_path);            res.add(tmp_path);            return;        }        if (path.get(beginWord) == null) return;        tmp_path.add(beginWord);        int nextDepth = (int) path.get(beginWord) - 1;        for (int i = 0; i < beginWord.length(); i++) {            char[] ca = beginWord.toCharArray();            for (char c = 'a'; c <= 'z'; c++) {                if (ca[i] == c) continue;                ca[i] = c;                String newWord = new String(ca);                if (path.get(newWord) != null && path.get(newWord) == nextDepth) {                    List<String> new_tmp_path = new ArrayList<>(tmp_path);                    dfs(newWord, endWord, set, new_tmp_path, res);                }            }        }    }        private void bfs(String start, String end, Set<String> set) {        Queue<String> queue = new LinkedList<>();        queue.add(start);        path.put(start, 0);        String current;        while (!queue.isEmpty()) {            current = queue.poll();            if (current == end) continue;            for (int i = 0; i < current.length(); i++) {                char[] ca = current.toCharArray();                for (char c = 'a'; c <= 'z'; c++) {                    if (ca[i] == c) continue;                    ca[i] = c;                    String newWord = new String(ca);                    if (newWord.equals(end) || set.contains(newWord)) {                        if (path.get(newWord) == null) {                            int depth = path.get(current);                            path.put(newWord, depth + 1);                            queue.add(newWord);                        }                    }                }            }        }    }