127. Word Ladder（BFS）

最新推荐文章于 2021-01-26 00:23:52 发布

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本文介绍了一种使用广度优先搜索(BFS)算法解决单词从开始到结束最短转换路径的问题，通过改变字母并确保每一步都在给定字典中找到有效单词。此算法能够高效地找出最短路径长度。

题目：

Given two words (beginWord and endWord), and a dictionary's word list, find the length of shortest transformation sequence frombeginWord 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.

Note:

Return 0 if there is no such transformation sequence.
All words have the same length.
All words contain only lowercase alphabetic characters.

求最短路径长度，相当于找字典树最小深度，用bfs

代码：

public class Solution {
    public int ladderLength(String beginWord, String endWord, Set<String> wordList) {
        Map<String, Integer> map = new HashMap<String, Integer>();
        LinkedList<String> queue = new LinkedList<String>();
        queue.add(beginWord);
        wordList.remove(beginWord);
        map.put(beginWord, 1);
        while(!queue.isEmpty())
        {
        	StringBuilder builder;
        	String cur = queue.pop();
        	int level = map.get(cur);
        	for(int i=0;i<cur.length();i++)
        	{
        		builder = new StringBuilder(cur);
        		for(int j='a';j<='z';j++)
        		{
        			builder.setCharAt(i, (char)j);
        			String temp = builder.toString();
        					
        			if(temp.equals(cur))
        			{
        				continue;
        			}
        			if(temp.equals(endWord))
        			{
        				return level+1;
        			}
        			if(wordList.contains(temp))
        			{
        				wordList.remove(temp);
        				queue.add(temp);
        				map.put(temp, level+1);
        			}
        		}
        	}
        }
        
        return 0;
    }
}