Given two words (beginWord and endWord), and a dictionary's word list, find all shortest transformation sequence(s) from beginWord to endWord, such that:
Only one letter can be changed at a timeEach 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); } } } } } }
