[LeetCode] Problem 127 - Word Ladder

Posted on | In ,

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:

  1. Only one letter can be changed at a time.
  2. 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

No.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.

No.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

1
2
3
4
5
6
7
8
9
10
11
12
13
14
15
16
17
18
19
20
21
22
23
24
25
26
27
28
29
30
31
32
33
34
35
36
37
public int ladderLength(String beginWord, String endWord, List<String> wordList) {
    int count = 1;
    Set<String> set = new HashSet<>();
    Queue<String> queue = new LinkedList<>();
    queue.offer(beginWord);

    for (String word : wordList)
        set.add(word);
    
    while (!queue.isEmpty()) {
        int size = queue.size();
        count++;

        for (int i = 0; i < size; i++) {
            String word = queue.poll();

            for (int idx = 0; idx < word.length(); idx++) {
                char[] newWord = word.toCharArray();

                for (char ch = 'a'; ch <= 'z'; ch++) {
                    newWord[idx] = ch;
                    String str = String.valueOf(newWord);

                    if (set.contains(str)) {
                        if (endWord.equals(str))
                            return count;

                        queue.offer(str);
                        set.remove(str);
                    }
                }
            }
        }
    }

    return 0;
}