LeetCode 126 Word Ladder II

来源：互联网发布：淘宝营销方式有哪些编辑：程序博客网时间：2024/06/06 06:37

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 time
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"]  ]

Note:

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

Runtime: 28 ms beats 99.16% of java submissions.

先用HashMap<String, ArrayList<String>> map 通过bfs广度优先，存下最短路径的形成过程，map类似于一个有起点和终点的图，然后dfs深度优先，寻找起点到终点的具体路径。

public List<List<String>> findLadders(String beginWord, String endWord, Set<String> wordList) {HashMap<String, ArrayList<String>> map = new HashMap();Set<String> start = new HashSet(), end = new HashSet();start.add(beginWord);end.add(endWord);BFS(start, end, wordList, map, true);List<List<String>> result = new ArrayList();List<String> cur = new ArrayList();cur.add(beginWord);DFS(beginWord, endWord, map, cur, result);return result;}private void BFS(Set<String> start, Set<String> end, Set<String> wordList, HashMap<String,ArrayList<String>> map, boolean forward) {if (start.size() > end.size()) {//为了提高效率 BFS(end, start, wordList, map, !forward);//forward 是正向，还是从end开始反向return;}wordList.removeAll(start);wordList.removeAll(end);boolean connected = false;Set<String> set = new HashSet();for (String s : start) {char[] chars = s.toCharArray();for (int i = 0, len = chars.length; i < len; i++) {char ch = chars[i];for (char x = 'a'; x <= 'z'; x++) {chars[i] = x;String word = new String(chars);if (end.contains(word) || (!connected && wordList.contains(word))) {if (end.contains(word)) connected = true;//找到最短的梯子，结束递归else set.add(word);if (forward) {ArrayList<String> cur = map.getOrDefault(s, new ArrayList<>());cur.add(word);map.put(s, cur);} else {ArrayList<String> cur = map.getOrDefault(word, new ArrayList<>());cur.add(s);map.put(word, cur);}}chars[i] = ch;}}}if (!connected && !set.isEmpty())BFS(set, end, wordList, map, forward);}private void DFS(String start, String end, HashMap<String, ArrayList<String>> map, List<String>cur, List<List<String>> result) {if (start.equals(end)) {result.add(new ArrayList(cur));//此处必须new ArrayList，因为cur是形参，会在递归中变化return;}if (!map.containsKey(start)) return;List<String> next = map.get(start);for (String i : next) {cur.add(i);DFS(i, end, map, cur, result);cur.remove(cur.size() - 1);}}

0 0