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:

1. Return an empty list if there is no such transformation sequence.

2. All words have the same length.

3. All words contain only lowercase alphabetic characters.

4. You may assume no duplicates in the word list.

5. You may assume beginWord and endWord are non-empty and are not the same.

* Method 1: dfs TLE

```Java

class Solution {

private int min = Integer.MAX_VALUE >> 1;

public List<List<String>> findLadders(String beginWord, String endWord, List<String> wordList) {

List<List<String>> result = new ArrayList<>();

Set<String> words = new HashSet<>(wordList);

if(!words.contains(endWord)) return result;

int len = beginWord.length();

List<String> temp = new ArrayList<String>();

temp.add(beginWord);

dfs(result, words, len, endWord, beginWord, temp);

return result;

}

private void dfs(List<List<String>> result, Set<String> words, int len, String endWord, String cur, List<String> temp){

if(cur.equals(endWord)){

if(temp.size() < min){

result.clear();

min = temp.size();

}

result.add(new ArrayList<String>(temp));

}else if(temp.size() < min){

for(int i = 0; i < len; i++){

char[] arr = cur.toCharArray();

for(char c = 'a'; c <= 'z'; c++){

arr[i] = c;

String next = new String(arr);

if(words.contains(next)){

words.remove(next);

temp.add(next);

dfs(result, words, len, endWord, next, temp);

temp.remove(temp.size() - 1);

words.add(next);

}

}

arr[i] = cur.charAt(i);

}

}

}

}

```

* Method 2: bfs AC

```Java

class Solution {

private Map<String, List<String>> map;

public List<List<String>> findLadders(String beginWord, String endWord, List<String> wordList) {

List<List<String>> result = new LinkedList<>();

Set<String> words = new HashSet<>(wordList);

if(!words.contains(endWord)) return result;

words.remove(beginWord);

LinkedList<String> queue = new LinkedList<>();

map = new HashMap<>();

int len = beginWord.length();

boolean found = false;

queue.add(beginWord);

while(!queue.isEmpty() && !found){

int size = queue.size();

Set<String> curLevel = new HashSet<>();

for(int i = 0; i < size; i++){

String cur = queue.poll();

if(cur.equals(endWord)) found = true;

char[] arr = cur.toCharArray();

for(int l = 0; l < len; l++){

for(char c = 'a'; c <= 'z'; c++){

if(c == cur.charAt(l)) continue;

arr[l] = c;

String next = new String(arr);

if(words.contains(next) || curLevel.contains(next)){

map.put(cur, !map.containsKey(cur) ? new ArrayList<String>(): map.get(cur));

map.get(cur).add(next);

curLevel.add(next);

if(words.contains(next)) queue.offer(next);

words.remove(next);

}

}

arr[l] = cur.charAt(l);

}

}

}

List<String> temp = new LinkedList<>();

temp.add(beginWord);

dfs(result, temp, endWord, beginWord);

return result;

}

private void dfs(List<List<String>> result, List<String> temp, String endWord, String cur){

if(cur.equals(endWord)){

result.add(new ArrayList<>(temp));

}else{

if(!map.containsKey(cur)) return;

List<String> adj = map.get(cur);

for(String s: adj){

temp.add(s);

dfs(result, temp, endWord, s);

temp.remove(temp.size() - 1);

}

}

}

}

```

* Method 1: DFS TLE

```JAVA

class Solution {

public int ladderLength(String beginWord, String endWord, List<String> wordList) {

int len = beginWord.length();

int res = dfs(wordList, endWord, beginWord, len, new boolean[wordList.size()]);

return (res == (Integer.MAX_VALUE >> 1)) ? 0 : res;

}

private int dfs(List<String> wordList, String endWord, String cur, int len, boolean[] visited){

if(cur.equals(endWord)) return 1;

else{

int min = Integer.MAX_VALUE >> 1;

for(int i = 0; i < wordList.size(); i++){

if(visited[i] || !isNext(cur, wordList.get(i), len)) continue;

visited[i] = true;

min = Math.min(min, dfs(wordList, endWord, wordList.get(i), len, visited) + 1);

visited[i] = false;

}

return min;

}

}

private boolean isNext(String pre, String next, int len){

int count = 0;

for(int i = 0; i < len; i++){

if(next.charAt(i) == pre.charAt(i)) continue;

else count++;

if(count > 1) return false;

}

return count == 1;

}

}

```

* Method 2: BFS AC

```Java

class Solution {

public int ladderLength(String beginWord, String endWord, List<String> wordList) {

Set<String> remain = new HashSet<>(wordList);

LinkedList<String> queue = new LinkedList<>();

if(!remain.contains(endWord)) return 0;

queue.add(beginWord);

int res = 0, len = beginWord.length();

while(!queue.isEmpty()){

res++;

int size = queue.size();

for(int i = 0; i < size; i++){

String cur = queue.poll();

if(cur.equals(endWord)) return res;

char[] arr = cur.toCharArray();

for(int l = 0; l < len; l++){

char origin = arr[l];

for(char c = 'a'; c <= 'z'; c++){

if(c == origin) continue;

arr[l] = c;

String next = new String(arr);

if(remain.contains(next)){

remain.remove(next);

queue.add(next);

}

}

arr[l] = origin;

}

}

}

return 0;

}

}

```

* Method 3: Bidirectional bfs

```Java

class Solution {

public int ladderLength(String beginWord, String endWord, List<String> wordList) {

Set<String> dict = new HashSet<>(wordList);

if(!dict.contains(endWord)) return 0;

Set<String> head = new HashSet<>();

Set<String> tail = new HashSet<>();

head.add(beginWord);

tail.add(endWord);

int step = 0, len = endWord.length();

while(!head.isEmpty() && !tail.isEmpty()){

++step;

int headSize = head.size(), tailSize = tail.size();

Set<String> temp = null;

if(headSize > tailSize){

temp = head;

head = tail;

tail = temp;

}

Set<String> next = new HashSet<>();

for(String s: head){

char[] arr = s.toCharArray();

for(int i = 0; i < len; i++){

for(char c = 'a'; c <= 'z'; c++){

arr[i] = c;

String newWord = new String(arr);

if(tail.contains(newWord)) return step + 1;

if(dict.contains(newWord)){

next.add(newWord);

dict.remove(newWord);

}

}

arr[i] = s.charAt(i);

}

}

head = next;

}

return 0;

}

}

```

### Conclusion

* How to choose dfs and bfs

1. bfs is designed for shortest path question.

2. At one single stage, we could get the next stage and in every single step, the result is optimized, we can use bfs.

3. dfs can also solve bfs questions while it takes longer time.

4. dfs is used for recording all possible solutions(combination and permutation questions).

1. [LeetCode 127. Word Ladder](https://www.jianshu.com/p/753bd585d57e)

2. [花花酱 LeetCode 127. Word Ladder](http://zxi.mytechroad.com/blog/searching/127-word-ladder/)

3. [什么时候用深搜（dfs）什么时候用广搜（bfs）（转）](https://www.cnblogs.com/Annetree/p/7199358.html)