[leetcode]126. Word LadderII@Java解题报告

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本文介绍了一种使用广度优先搜索(BFS)解决LeetCode上Word Ladder II问题的方法，该问题要求找出从一个单词转换到另一个单词的最短路径，并详细解释了如何通过构建图来寻找所有可能的最短转换路径。

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https://leetcode.com/problems/word-ladder-ii/#/description

Given two words (beginWord and endWord), and a dictionary's word list, find all shortest transformation sequence(s) from beginWordto 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.

For example,

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

Return

  [
    ["hit","hot","dot","dog","cog"],
    ["hit","hot","lot","log","cog"]
  ]

Note:

Return an empty list 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.

UPDATE (2017/1/20):
The wordList parameter had been changed to a list of strings (instead of a set of strings). Please reload the code definition to get the latest changes.

package go.jacob.day727;

import java.util.ArrayList;
import java.util.HashMap;
import java.util.LinkedList;
import java.util.List;
import java.util.Map;
import java.util.Queue;

/**
 * 126. Word Ladder II
 * 思路：图bfs的应用
 * @author Jacob
 *
 */
public class Demo2 {
	// res为最终结果
	List<List<String>> res;
	Map<String, List<String>> map;
	
	public List<List<String>> findLadders(String beginWord, String endWord, List<String> wordList) {
		res = new ArrayList<List<String>>();
		if (wordList.size() == 0)
			return res;
		// 最小步数初始化为最大值
		int min = Integer.MAX_VALUE;
		Queue<String> queue = new LinkedList<String>();
		queue.add(beginWord);
		map = new HashMap<String, List<String>>();
		// ladder用来记录每个string出现的最小序号
		Map<String, Integer> ladder = new HashMap<String, Integer>();

		for (String string : wordList)
			// 初始化为最大值
			ladder.put(string, Integer.MAX_VALUE);
		ladder.put(beginWord, 0);
		wordList.add(endWord);
		while (!queue.isEmpty()) {
			String word = queue.poll();
			int step = ladder.get(word) + 1;
			// 如果大于最小步数，直接跳出循环
			if (step > min)
				break;

			for (int i = 0; i < word.length(); i++) {
				StringBuilder sb = new StringBuilder();
				for (char c = 'a'; c <= 'z'; c++) {
					sb.setCharAt(i, c);
					String new_word = sb.toString();
					if (ladder.containsKey(new_word)) {
						// 只有第一遍历到的new_word才会加入到queue,map，因为step是非递减的
						if (step > ladder.get(new_word))
							continue;
						else if (step < ladder.get(new_word)) {
							queue.add(new_word);
							ladder.put(new_word, step);
						} else
							;
						// map中只存放word的所有前一个元素
						if (map.containsKey(new_word))
							map.get(new_word).add(word);
						else {
							List<String> list = new LinkedList<String>();
							list.add(word);
							map.put(new_word, list);
						}
						if (new_word.equals(endWord))
							min = step;
					}
				}
			}
		}
		//只能用LinkedList，不能用ArrayList。原因见backTrace方法
		LinkedList<String> result = new LinkedList<String>();
		backTrace(endWord, beginWord, result);

		return res;
	}
	
	private void backTrace(String word, String start, List<String> list) {
		if (word.equals(start)) {
			list.add(0, start);
			res.add(new ArrayList<String>(list));
			list.remove(0);
			return;
		}
		//注意：这里的list是LinkedList类型，所以插入index=0的位置，其余元素全部后移
		//如果是ArrayList，会把0处的元素替换
		list.add(0, word);
		if (map.get(word) != null)
			for (String s : map.get(word))
				backTrace(s, start, list);
		list.remove(0);
	}

}