算法描述：

Given two words (beginWord and endWord), and a dictionary's word list, 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 transformed word must exist in the word list. Note that beginWord is not a transformed word.

Note:

• Return 0 if there is no such transformation sequence.
• All words have the same length.
• All words contain only lowercase alphabetic characters.
• You may assume no duplicates in the word list.
• You may assume beginWord and endWord are non-empty and are not the same.

Example 1:

Input:
beginWord = "hit",
endWord = "cog",
wordList = ["hot","dot","dog","lot","log","cog"]

Output: 5

Explanation: As one shortest transformation is "hit" -> "hot" -> "dot" -> "dog" -> "cog",
return its length 5.

Example 2:

Input:
beginWord = "hit"
endWord = "cog"
wordList = ["hot","dot","dog","lot","log"]

Output: 0

Explanation: The endWord "cog" is not in wordList, therefore no possible transformation.

解题思路：广度优先搜索。用一个队列存储可能的候选字符串。用set 构建字典，并去重。对于每一个字符串，逐一选替换每一位字符，如果发现字典中存在这样的字符，则将该字符串加入到队列中，并在字典中取出该字符串。

    int ladderLength(string beginWord, string endWord, vector<string>& wordList) {
        unordered_set<string> dict(wordList.begin(),wordList.end());
        if(dict.find(endWord)==dict.end()) return 0;
        
        queue<string> que;
        que.push(beginWord);
        int level = 0;
        
        while(!que.empty()){
            level++;
            
            int levelSize = que.size();
            for(int i=0; i < levelSize; i++){
                string word = que.front();
                que.pop();
                
                for(int j =0; j < beginWord.size(); j++){
                    char orginal_char = word[j];
                    for(char c = 'a'; c <= 'z'; c++){
                        word[j] = c;
                        if(word == endWord) return level+1;
                        if(dict.find(word)==dict.end()) continue;
                        que.push(word);
                        dict.erase(word);
                    }
                    word[j]=orginal_char;                    
                }
            }
        }
        return 0;
        
    }