[LeetCode]Word Ladder

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.

思考：BFS。

class Solution {
public:
	int ladderLength(string start, string end, unordered_set<string> &dict) {
		queue<pair<string,int>> q;
		unordered_set<string> visited;
		q.push(make_pair(start, 1));
		visited.insert(start);
		while (!q.empty())
		{
			string curStr = q.front().first;
			int curStep = q.front().second;
			q.pop();
			for (int i = 0; i < curStr.size(); ++i)
			{
				string tmp = curStr;
				for (int j = 0; j < 26; ++j)
				{
					tmp[i] = j+'a';
					if(tmp == end)
						return curStep+1;
					if(visited.find(tmp) == visited.end() && dict.find(tmp) != dict.end())
					{
						q.push(make_pair(tmp, curStep+1));
						visited.insert(tmp);
					}
				}
			}
		}
		return 0;
	}
};