Given two words (start and end), and a dictionary, find the length of shortest transformation sequence from start to end, such that:
- Only one letter can be changed at a time
- Each intermediate word must exist in the dictionary
For example,
Given:
start = "hit"
end = "cog"
dict = ["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.
class Solution { public: int ladderLength(std::string start, std::string end, std::unordered_set<std::string> &dict) { std::unordered_map<std::string, int> dis; std::queue<std::string> q; dis[start] = 1; q.push(start); while (!q.empty()) { std::string word = q.front(); q.pop(); if (word == end) break; for (int i = 0; i < word.size(); i++) { for (int j = 0; j < 26; j++) { std::string newWord = word; newWord[i] = 'a' + j; if (dict.count(newWord) > 0 && dis.count(newWord) == 0) { dis[newWord] = dis[word] + 1; q.push(newWord); } } } } if (dis.count(end) == 0) return 0; return dis[end]; } };