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.
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.
Example 1:
Input: beginWord = "hit", endWord = "cog", wordList = ["hot","dot","dog","lot","log","cog"] Output: [ ["hit","hot","dot","dog","cog"], ["hit","hot","lot","log","cog"] ]
Example 2:
Input: beginWord = "hit" endWord = "cog" wordList = ["hot","dot","dog","lot","log"] Output: [] Explanation: The endWord "cog" is not in wordList, therefore no possible transformation.
思路：BFS，注意保存BFS过程中的信息。
1.定义一个HashMap<String,List<String>> route 保存路径。（方向为value到key）
2.广搜每一次循环搜一层，所以代码中不用记录String和层数的映射了。
3.最后用DFS，在HashMap<String,List<String>> route 搜索路径。
4.定义mark标记已搜到的单词，在下一层不会再搜到，mark[i]==0表示还未搜到，mark[i]==1表示已搜到，但还在当前 层。mark[i]==2表示搜到，并已经到了下一层，已不会搜到。
5.因为第一版本的BFS时是遍历wordList中每个元素，再每个字符串相匹配，这样速度慢，每次匹配一个元素要O（L*m）的复杂度，L为字符串长度，m为wordList的字符串个数，AC后要600+ms。所以在第二版本中先把wordList存在一个HashMap中，在替换每个要匹配的String中的一个字符，再到HashMap中找，理论上每次匹配时间复杂度为O（26*L），AC后218ms。
第一版本：
public List<List<String>> findLadders(String beginWord, String endWord, List<String> wordList) {
HashMap<String,List<String>> route=new HashMap<>();//list to string
LinkedList<String> queue=new LinkedList<>();
List<List<String>> list=new ArrayList<>();
int[] mark=new int[wordList.size()];
boolean isEnd=false;//标记已找到
queue.add(beginWord);
while(!isEnd&&queue.size()!=0){
int len=queue.size();
while((len--)>0){
String str=queue.poll();
for(int i=0;i<wordList.size();i++){
if(mark[i]!=2&&connect(wordList.get(i),str)){
if(route.get(wordList.get(i))==null){
List<String> l=new ArrayList<>();
l.add(str);
route.put(wordList.get(i),l);
}else{
route.get(wordList.get(i)).add(str);
}
if(endWord.equals(wordList.get(i))){
isEnd=true;//找到就是最后一次循环
}else{
if(mark[i]!=1){//本次循环已经找到次数
queue.add(wordList.get(i));
}
}
mark[i]=1;
}
}
}
for(int i=0;i<mark.length;i++){
if(mark[i]==1){
mark[i]=2;
}
}
}
List<String> l=new ArrayList<>();
if(route.get(endWord)==null) return list;
l.add(endWord);
addString(route,list,l,endWord,beginWord);
return list;
}
public void addString(HashMap<String,List<String>> route,List<List<String>> list,List<String> l,String word,String start){
if(start.equals(word)){
list.add(new ArrayList(l));
return;
}
for(String s:route.get(word)){
l.add(0,s);
addString(route,list,l,s,start);
l.remove(0);
}
}
public boolean connect(String str1,String str2){
if(str1.equals(str2)||str1.length()!=str2.length()) return false;
int count=0;
for(int i=0;i<str1.length();i++){
if(str1.charAt(i)!=str2.charAt(i)){
count++;
if(count>1){
return false;
}
}
}
return true;
}
第二版本：
public List<List<String>> findLadders(String beginWord, String endWord, List<String> wordList) {
HashMap<String,List<String>> route=new HashMap<>();//list to string
LinkedList<String> queue=new LinkedList<>();
List<List<String>> list=new ArrayList<>();
HashMap<String,Integer> map=new HashMap<>();//记录字符串和下标映射
for(int i=0;i<wordList.size();i++){
map.put(wordList.get(i),i);
}
int[] mark=new int[wordList.size()];
boolean isEnd=false;//标记已找到
queue.add(beginWord);
while(!isEnd&&queue.size()!=0){
int len=queue.size();
while((len--)>0){
String str=queue.poll();
for(int i=0;i<str.length();i++){
for(char c='a';c<='z';c++){
if(str.charAt(i)!=c){
StringBuilder sb=new StringBuilder(str);
sb.setCharAt(i,c);
if(map.get(sb.toString())!=null){//在dict中找到了string
int index=map.get(sb.toString());
if(mark[index]==2){
continue;
}
if(route.get(sb.toString())==null){//路径str到sb.toString()
List<String> l=new ArrayList<>();
l.add(str);
route.put(sb.toString(),l);//key存下一个
}else{
route.get(sb.toString()).add(str);
}
if(endWord.equals(sb.toString())){
isEnd=true;//找到就是最后一次循环
}else{
if(mark[index]!=1){//本次循环已经找到次数
queue.add(sb.toString());
}
}
mark[index]=1;
}
}
}
}
}
for(int i=0;i<mark.length;i++){
if(mark[i]==1){
mark[i]=2;
}
}
}
List<String> l=new ArrayList<>();
if(route.get(endWord)==null) return list;
l.add(endWord);
addString(route,list,l,endWord,beginWord);
return list;
}
public void addString(HashMap<String,List<String>> route,List<List<String>> list,List<String> l,String word,String start){
if(start.equals(word)){
list.add(new ArrayList(l));
return;
}
for(String s:route.get(word)){
l.add(0,s);
addString(route,list,l,s,start);
l.remove(0);
}
}
}