算法细节系列（20）：Word Ladder系列
算法细节系列（20）：Word Ladder系列
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发布于 2019-05-26 09:46:12
发布于 2019-05-26 09:46:12
版权声明：本文为博主原创文章，未经博主允许不得转载。 https://cloud.tencent.com/developer/article/1434754
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：
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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原始发表：2017年05月18日，如有侵权请联系 cloudcommunity@tencent.com 删除
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