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## 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. 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. Example 1:

Input:
beginWord = "hit",
endWord = "cog",
wordList = ["hot","dot","dog","lot","log","cog"]

Output: 5

Explanation: As one shortest transformation is "hit" -> "hot" -> "dot" -> "dog" -> "cog",
return its length 5.


Example 2:

Input:
beginWord = "hit"
endWord = "cog"
wordList = ["hot","dot","dog","lot","log"]

Output: 0

Explanation: The endWord "cog" is not in wordList, therefore no possible transformation.


## Code

class Solution {
public int ladderLength(String beginWord, String endWord, List<String> wordList) {
Set<String> set = new HashSet<>(wordList);

queue.offer(beginWord);

int level = 1;
while(!queue.isEmpty()){
int size = queue.size();
for(int i = 0; i < size; i++){
String curr = queue.poll();
for(int j = 0; j < curr.length(); j++){
char[] arr = curr.toCharArray();
for(char x = 'a'; x <= 'z'; x++){
arr[j] = x;
String temp = new String(arr);
if(set.contains(temp)){
if(temp.equals(endWord)) return level + 1;
queue.offer(temp);
set.remove(temp);
}
}
}
}

level++;
}

return 0;
}
}

• time: O(N * (26 ^ L)), L = len(word), N = len(wordList)
• space: O(N)

bidirectional BFS

class Solution {
public int ladderLength(String beginWord, String endWord, List<String> wordList) {
Set<String> dict = new HashSet<>();
for (String word : wordList) dict.add(word);

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

Set<String> q1 = new HashSet<>();
Set<String> q2 = new HashSet<>();

int l = beginWord.length();
int steps = 0;
// 每次从左边扩展一步或者从右边扩展一步
while (!q1.isEmpty() && !q2.isEmpty()) {
steps++;

// 交替更换两个queue
if (q1.size() > q2.size()) {
Set<String> tmp = q1;
q1 = q2;
q2 = tmp;
}

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

for (String w : q1) {
char[] chs = w.toCharArray();
for (int i = 0; i < l; ++i) {
char ch = chs[i];
for (char c = 'a'; c <= 'z'; ++c) {
chs[i] = c;
String t = new String(chs);
// 在另外一个queue中碰到了, 说明找到了路径
if (q2.contains(t)) return steps + 1;
if (!dict.contains(t)) continue;
dict.remove(t);