127. Word Ladder (很重要！！！)

最新推荐文章于 2022-05-20 22:03:23 发布

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最新推荐文章于 2022-05-20 22:03:23 发布

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本文介绍了一种算法，用于寻找两个单词间最短的转换路径，每次仅改变一个字母，并确保每一步骤中的单词都存在于给定的词典中。通过使用广度优先搜索策略，可以在最短步数内完成从起始单词到目标单词的转变。

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

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class Solution {
public:
	int ladderLength(string beginWord, string endWord, unordered_set<string>& wordList) {
		queue<string> Q;
		set<string>visited;
		Q.push(beginWord);
		visited.insert(beginWord);
		int cnt = 1;
		while (!Q.empty()){
			cnt++;
			int size = Q.size();
			while (size--){
				string front = Q.front();
				Q.pop();
				for (int i = 0; i < front.size(); i++){
					for (char ch = 'a'; ch <= 'z'; ch++){
						if (ch == front[i]) continue;
						string tmp = front;
						tmp[i] = ch;
						if (wordList.count(tmp) && !visited.count(tmp)){
							if (tmp == endWord) return cnt;
							Q.push(tmp);
							visited.insert(tmp);
						}
					}
				}
			}
		}
		return 0;
	}
};