127. Word Ladder (很重要！！！)

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

1. Only one letter can be changed at a time
2. 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;}};