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.

class Solution {

public:

    int ladderLength(string start, string end, unordered_set<string> &dict) {

        if (start == end) {

            return ;

        }

        int n = start.size();

        if (n <  || n != end.size()) {

            return ;

        }

        queue<string> q;

        q.push(start);

        dict.erase(start);

        int length = ;

        while (!q.empty()) {

            int size = q.size();

            for (int i = ; i < size; ++i) {

                string cur = q.front();

                q.pop();

                for (int j = ; j < n; ++j) {

                    char oldChar = cur[j];

                    for (char c = 'a'; c <= 'z'; ++c) {

                        if (c == oldChar) {

                            continue;

                        }

                        cur[j] = c;

                        if (cur == end) {

                            return length;

                        }

                        if (dict.find(cur) != dict.end()) {

                            q.push(cur);

                            dict.erase(cur);

                        }

                    }

                    cur[j] = oldChar;

                }

            }

            ++length;

        }

        return ;

    }

};