LeetCode-127-Word Ladder

最新推荐文章于 2025-08-15 16:44:13 发布

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最新推荐文章于 2025-08-15 16:44:13 发布

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文章标签：数据结构与算法

原文链接：http://www.cnblogs.com/nobodywang/p/10352994.html

本文介绍了一种使用广度优先搜索算法解决单词转换问题的方法，即从开始单词通过替换一个字母逐步转换到目标单词，每一步转换后的单词必须在给定的字典中。文章详细解释了算法实现，包括如何利用队列存储候选字符串和如何使用哈希集合进行字典构建和去重。

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算法描述：

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.

解题思路：广度优先搜索。用一个队列存储可能的候选字符串。用set 构建字典，并去重。对于每一个字符串，逐一选替换每一位字符，如果发现字典中存在这样的字符，则将该字符串加入到队列中，并在字典中取出该字符串。

    int ladderLength(string beginWord, string endWord, vector<string>& wordList) {
        unordered_set<string> dict(wordList.begin(),wordList.end());
        if(dict.find(endWord)==dict.end()) return 0;
        
        queue<string> que;
        que.push(beginWord);
        int level = 0;
        
        while(!que.empty()){
            level++;
            
            int levelSize = que.size();
            for(int i=0; i < levelSize; i++){
                string word = que.front();
                que.pop();
                
                for(int j =0; j < beginWord.size(); j++){
                    char orginal_char = word[j];
                    for(char c = 'a'; c <= 'z'; c++){
                        word[j] = c;
                        if(word == endWord) return level+1;
                        if(dict.find(word)==dict.end()) continue;
                        que.push(word);
                        dict.erase(word);
                    }
                    word[j]=orginal_char;                    
                }
            }
        }
        return 0;
        
    }