[leetcode]126. Word LadderII@Java解题报告

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:

  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.

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);
	}

}




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