Word Ladder**

Given two words (beginWord and endWord), and a dictionary, find the length of shortest transformation sequence from beginWord to endWord, 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.

Analyse: Refered to this. 

Runtime: 276ms. 

class Solution {
public:
    int ladderLength(string beginWord, string endWord, unordered_set<string>& wordDict) {
        wordDict.insert(endWord); //consider the string set as a graph
        queue<pair<string, int> > q;
        q.push(make_pair(beginWord, 1)); //start from the beginWord
        
        while(!q.empty()){
            string s = q.front().first; //string to be compared
            int result = q.front().second; //the length of the ladder
            
            if(s == endWord) return result; //reach to the last string
            q.pop();
            vector<string> neighbours = findNeighbours(s, wordDict); //find all neighbour string to the s
            for(int i = 0; i < neighbours.size(); i++)
                q.push(make_pair(neighbours[i], result + 1));
        }
        return 0;
    }
    
    vector<string> findNeighbours(string s, unordered_set<string>& wordDict){
        vector<string> result;
        for(int i = 0; i < s.length(); i++){
            char temp = s[i];
            for(int j = 0; j < 26; j++){// O(26 * s.length())
                if(temp == 'a' + j) continue; //if the character is the same, then move to the next character
                s[i] = 'a' + j;
                if(wordDict.find(s) != wordDict.end()){
                    result.push_back(s);
                    wordDict.erase(s);
                }
            }
            s[i] = temp;
        }
        return result;
    }
};