算法细节系列（20）：Word Ladder系列

详细代码可以fork下Github上leetcode项目，不定期更新。

题目摘自leetcode：
1. Leetcode 127: Word Ladder
2. Leetcode 126: Word Ladder II

Leetcode 127: Word Ladder

Problem:

Given two words (beginWord and endWord), and a dictionary’s word list, find the length of shortest transformation sequence from beginWord to endWord, such that:

• Only one letter can be changed at a time.
• Each transformed word must exist in the word list. Note that beginWord is not a transformed word.

Example:

Given:
beginWord = “hit”
endWord = “cog”
wordList = [“hot”,”dot”,”dog”,”lot”,”log”,”cog”]
As one shortest transformation is “hit” -> “hot” -> “dot” -> “dog” -> “cog”,
return its length 5.

Note:

• Return 0 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.

这道题其实不难，但要想到这种解法却要费一番周折，如果对最短路径搜索熟悉的话，相信你一眼就能看出答案了，并且我们要论证一点，为什么最短路径算法对这道题来说是正确解法。

我的思路：
DFS，把所有编辑距离为1的单词连接在一块，构建一个MAP（邻接矩阵）。这样之后，我们就可以从beginWord开始DFS搜索了，中间需要状态记录。代码如下：

public int ladderLength(String beginWord, String endWord, List<String> wordList) {Map<String, List<String>> map = new HashMap<>();map.put(beginWord, new ArrayList<>());for (String word : wordList){map.put(word, new ArrayList<>());}for (String key : map.keySet()){List<String> container = map.get(key);for (String word : wordList){if (oneDiff(key, word)){container.add(word);}}map.put(key, container);}int ans = helper(map, beginWord, endWord, new HashSet<>());return ans >= 1 << 30 ? 0 : ans;}private int helper(Map<String, List<String>> map,String beginWord, String endWord, Set<String> visited){if (visited.contains(beginWord)) return 1<<30;visited.add(beginWord);for (String find : map.get(beginWord)){if (find.equals(endWord)){visited.remove(beginWord);return 2;}}int min = Integer.MAX_VALUE;for (String find : map.get(beginWord)){int x = 1 + helper(map, find, endWord, visited);min = Math.min(min, x);}visited.remove(beginWord);return min;}private boolean oneDiff(String a, String b){if (a.equals(b)) return false;char[] aa = a.toCharArray();char[] bb = b.toCharArray();int oneDiff = 0;for (int i = 0; i < aa.length; i++){if (aa[i] != bb[i]){oneDiff ++;if (oneDiff >= 2) return false;}}return true;}

代码没有多大问题，典型的DFS+状态回溯，遍历搜索每一条到达endWord的路径，找寻最短路径。但很可惜TLE了，直观上来看是因为为了拿到到endWord的最短路径，我们需要遍历每一条到endWord的路径，这是递归求解的一个特点。但实际情况，我们可以省去某些点的遍历。

就那这个问题来说，如从beginWord开始搜索，如

beginWord = "hit" endWord = "cog" wordList = ["hot","dot","dog","lot","log","cog"] wordList中编辑距离为1的单词有： a. hot 此时BFS搜索与"hot"最近距离的单词，有： a. dot b. lot 再BFS搜索"dot"时，有： a. cog 所以我们只需要BFS三次就能得到正确答案，而DFS中，需要DFS至少三次。

上述过程就是经典的Dijkstra算法，代码如下：

public int ladderLength(String beginWord, String endWord, List<String> wordList) {List<String> reached = new ArrayList<>();reached.add(beginWord);Set<String> wordSet = new HashSet<>(wordList);if(!wordSet.contains(endWord)) return 0;wordSet.add(endWord);int distance = 1;while (!reached.contains(endWord)){ //到达该目的地List<String> toAdd = new ArrayList<>();for (String each : reached){for (int i = 0; i < each.length(); i++){char[] chars = each.toCharArray();for (char c = 'a'; c <= 'z'; c++){chars[i] = c;String wd = new String(chars);if (wordSet.contains(wd)){toAdd.add(wd);wordSet.remove(wd); //记录访问状态}}}}distance ++;if (toAdd.size() == 0) return 0; //没有编辑距离为1的单词reached = toAdd;}return distance;}

Leetcode 126: Word Ladder II

Problem:

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

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.

这道题的思路让我对DFS和BFS有了一些基本理解，但还不够深刻，咋说呢，我没想到BFS和DFS还可以分工合作，BFS用来快速求出最小distance，而DFS则用来遍历所有路径，两种遍历方法各有长处，综合起来就能解决该问题了，所以我写了一个版本，代码如下：

public List<List<String>> findLadders(String beginWord, String endWord, List<String> wordList) {Map<String, List<String>> map = new HashMap<>();map.put(beginWord, new ArrayList<>());for (String word : wordList){map.put(word, new ArrayList<>());}for (String key : map.keySet()){List<String> container = map.get(key);for (String word : wordList){if (oneDiff(key, word)){container.add(word);}}map.put(key, container);}int distance = bfs(beginWord, endWord, wordList);List<List<String>> ans = new ArrayList<>();dfs(map, beginWord, endWord, ans, new ArrayList<>(), distance);return ans;}private void dfs(Map<String, List<String>> map,String beginWord, String endWord, List<List<String>> ans, List<String> path, int distance){path.add(beginWord);if (distance == 0){path.remove(path.size()-1); return;}if (beginWord.equals(endWord)){ans.add(new ArrayList<>(path));path.remove(path.size()-1);return;}for (String find : map.get(beginWord)){dfs(map, find, endWord, ans, path, distance-1);}path.remove(path.size()-1);}private int bfs(String beginWord, String endWord, List<String> wordList) {List<String> reached = new ArrayList<>();reached.add(beginWord);Set<String> wordSet = new HashSet<>(wordList);if(!wordSet.contains(endWord)) return 0;wordSet.add(endWord);int distance = 1;while (!reached.contains(endWord)){ //达到该目的地List<String> toAdd = new ArrayList<>();for (String each : reached){for (int i = 0; i < each.length(); i++){char[] chars = each.toCharArray();for (char c = 'a'; c <= 'z'; c++){chars[i] = c;String wd = new String(chars);if (wordSet.contains(wd)){toAdd.add(wd);wordSet.remove(wd);}}}}distance ++;if (toAdd.size() == 0) return 0;reached = toAdd;}return distance;}private boolean oneDiff(String a, String b){if (a.equals(b)) return false;char[] aa = a.toCharArray();char[] bb = b.toCharArray();int oneDiff = 0;for (int i = 0; i < aa.length; i++){if (aa[i] != bb[i]){oneDiff ++;if (oneDiff >= 2) return false;}}return true;}

思路相当清楚了，以为能够AC，结果发现TLE了，说明该题对时间的要求很高，从上述代码我们也能发现一些基本问题，如BFS遍历时可以构建MAP，而不用单独构建MAP，非常耗时。其次，最关键的问题在于DFS，此版本的DFS没有进行剪枝处理，剪枝能省去很多时间，所以我还需要对BFS进行改进。

思路：
首先，我们来看看上述代码构建图的一个模型，如下图所示：

很明显，如果我们对BFS没有做任何限制，我们拿到的邻接表一定是上述探头斯，而此时如果用DFS进行搜索时，如从“hot”开始，它会搜索：

一条可能的搜索路径： hot ---> dot ---> dog ---> cog 但与此同时DFS还会搜索路径： hot ---> dot ---> tot ---> hot 上述路径很明显不需要DFS，但因为边的相连，使得这种没必要的搜索也将继续。

所以一个优化点就在于，好马不吃回头草，存在环路的回头草绝对不是达到endWord的最短路径。很遗憾，邻接表无法表示这种非环的图，所以想法就是用一个Map<String,Integer>来记录到达每个单词的最短路径，一旦map中有该单词，就不再更新最短路径（避免环路搜索）

所以BFS代码如下：

private int bfs(String beginWord, String endWord, Set<String> wordDict, Map<String, Integer> distanceMap,Map<String, List<String>> map) {if (!wordDict.contains(endWord))return 0;map.put(beginWord, new ArrayList<>());for (String word : wordDict) {map.put(word, new ArrayList<>());}Queue<String> queue = new LinkedList<>();queue.offer(beginWord);distanceMap.put(beginWord, 1);while (!queue.isEmpty()) {int count = queue.size();boolean foundEnd = false;// 这种循环遍历很有意思，看作一个整体for (int i = 0; i < count; i++) {String cur = queue.poll();int curDistance = distanceMap.get(cur);List<String> neighbors = getNeighbors(cur, wordDict);if (neighbors.size() == 0)return 0;for (String neighbor : neighbors) {map.get(cur).add(neighbor);//存在环的情况，不去更新最短路径if (!distanceMap.containsKey(neighbor)) {distanceMap.put(neighbor, curDistance + 1);if (endWord.equals(neighbor)) {foundEnd = true;} else {queue.offer(neighbor);}}}}//一旦抵到了endWord，我们就放弃建立后续的图if (foundEnd)break;}return distanceMap.get(endWord);}

上述代码在BFS时，与endWord无关的那些结点都丢弃掉了，且解决了有环路的情况。图结构如下所示：

这样在DFS构建路径时，它的速度就比原先要快得多。在BFS中还需要注意一个函数【getNeighbors()】，刚开始我写的这版程序也超时了，苦思许久都找不到原因，后来才发现是getNeighbors的玄机，它在建立邻接表时，一定要使用【HashSet】的搜索方法，而不要用原生的【List】的搜索方法。

所以完整代码如下：

public List<List<String>> findLadders(String beginWord, String endWord, List<String> wordList) {Map<String, List<String>> map = new HashMap<>();Map<String, Integer> distanceMap = new HashMap<>();Set<String> wordDict = new HashSet<>(wordList);wordDict.add(beginWord);int distance = bfs(beginWord, endWord, wordDict, distanceMap, map);List<List<String>> ans = new ArrayList<>();if (distance == 0)return ans;dfs(map, beginWord, endWord, ans, new ArrayList<>(), distance, distanceMap);return ans;}private void dfs(Map<String, List<String>> map, String beginWord, String endWord, List<List<String>> ans,List<String> path, int distance, Map<String, Integer> distanceMap) {path.add(beginWord);if (distance == 0) {path.remove(path.size() - 1);return;}if (beginWord.equals(endWord)) {ans.add(new ArrayList<>(path));path.remove(path.size() - 1);return;}for (String find : map.get(beginWord)) {if (!distanceMap.containsKey(find))continue;if (distanceMap.get(beginWord) + 1 == distanceMap.get(find))dfs(map, find, endWord, ans, path, distance - 1, distanceMap);}path.remove(path.size() - 1);}private int bfs(String beginWord, String endWord, Set<String> wordDict, Map<String, Integer> distanceMap,Map<String, List<String>> map) {if (!wordDict.contains(endWord))return 0;map.put(beginWord, new ArrayList<>());for (String word : wordDict) {map.put(word, new ArrayList<>());}Queue<String> queue = new LinkedList<>();queue.offer(beginWord);distanceMap.put(beginWord, 1);while (!queue.isEmpty()) {int count = queue.size();boolean foundEnd = false;for (int i = 0; i < count; i++) {String cur = queue.poll();int curDistance = distanceMap.get(cur);List<String> neighbors = getNeighbors(cur, wordDict);if (neighbors.size() == 0)return 0;for (String neighbor : neighbors) {map.get(cur).add(neighbor);if (!distanceMap.containsKey(neighbor)) {distanceMap.put(neighbor, curDistance + 1);if (endWord.equals(neighbor)) {foundEnd = true;} else {queue.offer(neighbor);}}}}if (foundEnd)break;}return distanceMap.get(endWord);}private List<String> getNeighbors(String word, Set<String> wordList) {List<String> ans = new ArrayList<>();for (int i = 0; i < word.length(); i++) {char[] cc = word.toCharArray();for (char c = 'a'; c <= 'z'; c++) {cc[i] = c;String newWord = new String(cc);if (wordList.contains(newWord)) {if (newWord.equals(word))continue;ans.add(newWord);}}}return ans;}

DFS是一个典型的回溯+剪枝的递归方法，凡是函数返回的地方，我们都需要进行状态还原，注意再注意。

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