算法: 词梯子127. Word Ladder

Posted 2021-12-03 架构师易筋

127. Word Ladder

A transformation sequence from word beginWord to word endWord using a dictionary wordList is a sequence of words beginWord -> s1 -> s2 -> … -> sk such that:

• Every adjacent pair of words differs by a single letter.
• Every si for 1 <= i <= k is in wordList. Note that beginWord does not need to be in wordList.
• sk == endWord

Given two words, beginWord and endWord, and a dictionary wordList, return the number of words in the shortest transformation sequence from beginWord to endWord, or 0 if no such sequence exists.

Example 1:

Input: beginWord = "hit", endWord = "cog", wordList = ["hot","dot","dog","lot","log","cog"]
Output: 5
Explanation: One shortest transformation sequence is "hit" -> "hot" -> "dot" -> "dog" -> cog", which is 5 words long.

Example 2:

Input: beginWord = "hit", endWord = "cog", wordList = ["hot","dot","dog","lot","log"]
Output: 0
Explanation: The endWord "cog" is not in wordList, therefore there is no valid transformation sequence.

Constraints:

• 1 <= beginWord.length <= 10
• endWord.length == beginWord.length
• 1 <= wordList.length <= 5000
• wordList[i].length == beginWord.length
• beginWord, endWord, and wordList[i] consist of lowercase English letters.
• beginWord != endWord
• All the words in wordList are unique.

广度优先算法

广度优先算法，想象一下类似wave波浪一圈一圈往外扩展，问题迎刃而解。

class Solution {
    public int ladderLength(String beginWord, String endWord, List<String> wordList) {
        Set<String> set = new HashSet<>();
        for (String s: wordList) set.add(s);
        int count = 1;
        set.remove(beginWord);
        Deque<String> deque = new LinkedList<>();
        deque.offer(beginWord);
        while (!deque.isEmpty()) {
            int size = deque.size();
            count++;
            for (int i = 0; i < size; i++) {
                char[] chars = deque.pollFirst().toCharArray();
                for (int k = 0; k < chars.length; k++) {
                   char c = chars[k];
                   for (char rc = 'a'; rc <= 'z'; rc++) {
                        if (rc == c) continue;
                        chars[k] = rc;
                        String news = String.valueOf(chars);
                        if (set.remove(news)) {
                            if (news.equals(endWord)) return count;
                            deque.offer(news);
                        }
                        chars[k] = c;
                    }
                }
            }
        }
        
        return 0;
    }
}