[LeetCode] Word Ladder

[Problem]

Given two words (start and end), and a dictionary, find the length of shortest transformation sequence from start to end, such that:

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

[Analysis]

深度优先搜索

[Solution]

class Solution {
public:
	int ladderLength(string start, string end, unordered_set<string> &dict) {
		// Start typing your C/C++ solution below
		// DO NOT write int main() function
		if(start == end){
			return 1;
		}

		// initial
		int cnt = 1, child = 0, res = 0;
		queue<string> myQueue;
		set<string> used;
		myQueue.push(start);
		used.insert(start);
		if(dict.find(end) == dict.end())
			dict.insert(end);

		// BFS
		while(!myQueue.empty()){
			string word = myQueue.front();
			myQueue.pop();
			cnt--;
			if(cnt == 0){
				res++;
			}

			// match
			if(word == end){
				if(cnt > 0)res++;
				return res;
			}

			// DFS
			for(int i = 0; i < word.length(); ++i){
				string tmp = word;
				for(int j = 0; j < 26; ++j){
					tmp[i] = char('a' + j);
					if(dict.find(tmp) != dict.end() && used.find(tmp) == used.end()){
						myQueue.push(tmp);
						used.insert(tmp);
						child++;
					}
				}
			}
			// next level
			if(cnt == 0){
				cnt = child;
				child = 0;
			}
		}
		return 0;
	}
};

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