LeetCode Word Ladder

最新推荐文章于 2025-08-16 00:47:01 发布

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本文介绍了如何通过BFS算法找到从一个单词转换为另一个单词的最短序列，序列中每个单词只能改变一个字母且必须存在于给定的词汇列表中。以hit到cog为例，展示了从hit到cog的最短转换路径长度为5。

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Description:

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 intermediate word must exist in the word list

For example,

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

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

Solution:

第一反应就是最短路，dijkstra的做法就是BFS

这里有个小trick，为了省下使用visited数字，每次进入队列的String都直接去掉

<span style="font-size:18px;">import java.util.*;

public class Solution {
	public int ladderLength(String beginWord, String endWord,
			Set<String> wordList) {

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

		queue.add(beginWord);
		wordList.remove(beginWord);
		int n = beginWord.length();
		int step = 0;
		String cur, neo;
		char[] cur_char = new char[n];
		while (!queue.isEmpty()) {
			int size = queue.size();
			step++;
			for (int i = 0; i < size; i++) {
				cur = queue.poll();
				cur_char = cur.toCharArray();
				for (int j = 0; j < n; j++) {
					for (int c = 'a'; c <= 'z'; c++) {
						if (cur_char[j] == c)
							continue;
						cur_char[j] = (char) c;
						neo = new String(cur_char);
						if (neo.equals(endWord))
							return step + 1;
						if (wordList.contains(neo)) {
							queue.add(neo);
							wordList.remove(neo);
						}
						cur_char[j] = cur.charAt(j);
					}
				}
			}
		}

		return 0;
	}
}</span>