leetcode 127. Word Ladder

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本文介绍了一种使用广度优先搜索(BFS)算法解决单词转换问题的方法，旨在找到从开始单词到目标单词的最短转换路径。通过将单词视为图中的节点，每一步转换视为边，利用BFS遍历图的特性，可以有效地找到最短路径。

题目

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 transformed word must exist in the word list. Note that beginWord is not a transformed word.

Note:

Return 0 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.

Example 1:

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

Output: 5

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

Example 2:

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

Output: 0

Explanation: The endWord "cog" is not in wordList, therefore no possible transformation.

题意：

给定一个开始字串，给定一个结尾字串，在给定一个词典，我们需要通过这个词典，将开始字串转换为结尾字串

每次只能转变一个字符，这个字符必须在词典里

问最少需要多少步转换

无法转换的返回0

解答

这题我使用BFS来进行解答，我们需要有两个集，一个是已经访问过的词典集reach，一个是未访问的（也就是wordList）

我们先将beginWord加入访问过的词典集reach。之后每次都执行这样的步骤：

寻找reach集可以变化位的词（也就是reach集中词改变一个字母可以得到的词），将词加入reach集，并从wordList中删除

不断执行，直到找到endWord，中间reach集变换了几次，就是需要几次转化。

以题目示例1说明

step 1:

reach={"hit"} wordLIist={"hot","dot","dog","lot","log","cog"}

step2

reach={"hot"} wordLIist={"dot","dog","lot","log","cog"}

step3

reach={"dot","lot"} wordLIist={"dog","log","cog"}

step4

reach={"dog","log"} wordLIist={"cog"}

step5

reach={"cog"} wordLIist={}

遇到的一个小问题：超时 TLE

经过研究，ArrayList调用 contains 复杂度是O（n），所以需要用一个更好的数据结构去代替，我使用了 Hashmap

成功AC。

代码：

class Solution {
    public int ladderLength(String beginWord, String endWord, List<String> wordList) {
        
        HashSet<String> dict = new HashSet<>();     //用hashmap来替代list，降低查找复杂度
        for(String word : wordList) {
           dict.add(word);
        }
        if(endWord.contains(endWord)==false) return 0;
        int distance=1;
        List<String> reach=new ArrayList<String>();
        reach.add(beginWord);
        while(true){
            if(reach.contains(endWord)) return distance;
            int bound=reach.size();
            for(int j=0;j<bound;j++){
                
                for(int x=0;x<beginWord.length();x++){
                    char []temp=reach.get(0).toCharArray();
                    for(char y='a';y<='z';y++){
                        temp[x]=y;
                        String s=String.valueOf(temp);
                        if(dict.contains(s)){
                            reach.add(s);
                            dict.remove(s);
                        }
                    }
                }
                reach.remove(0);
            }
            distance++;
            if(reach.size()==0) return 0;
        }
    }
}