### 描述

Given two words (start and end), and a dictionary, find the length of shortest transformation sequence from start to end, such that:

• Only one letter can be changed at a time
• Each intermediate word must exist in the dictionary

For example, Given:

start = "hit"
end = "cog"
dict = ["hot","dot","dog","lot","log"]


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.

### 单队列

// Word Ladder
// 时间复杂度O(n)，空间复杂度O(n)
public class Solution {
public int ladderLength(String beginWord, String endWord, Set<String> wordList) {
HashSet<State> visited = new HashSet<>(); // 判重

final Function<State, Boolean> stateIsValid = (State s) ->
wordList.contains(s.word) || s.word.equals(endWord);
final Function<State, Boolean> stateIsTarget = (State s) ->
s.word.equals(endWord);

final Function<State, HashSet<State> > stateExtend = (State s) -> {
HashSet<State> result = new HashSet<>();

char[] array = s.word.toCharArray();
for (int i = 0; i < array.length; ++i) {
final char old = array[i];
for (char c = 'a'; c <= 'z'; c++) {
// 防止同字母替换
if (c == array[i]) continue;

array[i] = c;
State newState = new State(new String(array), s.level+1);

if (stateIsValid.apply(newState) &&
!visited.contains(newState)) {
}
array[i] = old; // 恢复该单词
}
}

return result;
};

State startState = new State(beginWord, 0);
q.offer(startState);
while (!q.isEmpty()) {
State state = q.poll();

if (stateIsTarget.apply(state)) {
return state.level + 1;
}

HashSet<State> newStates = stateExtend.apply(state);
for (State newState : newStates) {
q.offer(newState);
}
}
return 0;
}

static class State {
String word;
int level;

public State(String word, int level) {
this.word = word;
this.level = level;
}

@Override
public int hashCode() {
return word.hashCode();
}

@Override
public boolean equals(Object other) {
if (this == other) return true;
if (this.hashCode() != other.hashCode()) return false;
if (!(other instanceof State)) return false;

return this.word.equals(((State) other).word);
}
}
}


### 双队列

// Word Ladder
// 时间复杂度O(n)，空间复杂度O(n)
public class Solution {
public int ladderLength(String beginWord, String endWord, Set<String> wordList) {
Queue<String> current = new LinkedList<>(); // 当前层
Queue<String> next = new LinkedList<>();    // 下一层
HashSet<String> visited = new HashSet<>();  // 判重

int level = -1;  // 层次

final Function<String, Boolean> stateIsValid = (String s) ->
wordList.contains(s) || s.equals(endWord);
final Function<String, Boolean> stateIsTarget = (String s) ->
s.equals(endWord);

final Function<String, HashSet<String> > stateExtend = (String s) -> {
HashSet<String> result = new HashSet<>();

char[] array = s.toCharArray();
for (int i = 0; i < array.length; ++i) {
final char old = array[i];
for (char c = 'a'; c <= 'z'; c++) {
// 防止同字母替换
if (c == array[i]) continue;

array[i] = c;
String newState = new String(array);

if (stateIsValid.apply(newState) &&
!visited.contains(newState)) {
}
array[i] = old; // 恢复该单词
}
}

return result;
};

current.offer(beginWord);
while (!current.isEmpty()) {
++level;
while (!current.isEmpty()) {
// 千万不能用 const auto&，pop() 会删除元素，
// 引用就变成了悬空引用
String state = current.poll();

if (stateIsTarget.apply(state)) {
return level + 1;
}

HashSet<String> newStates = stateExtend.apply(state);
for (String newState : newStates) {
next.offer(newState);
}
}
// swap
Queue<String> tmp = current;
current = next;
next = tmp;
}
return 0;
}
}