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.

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我发现我现在实力很弱，得更加努力了。

这个题是说的就是把一个字符串改变其中一个字符，并且保证改变后的字符串再给定的一个字典上有，然后，如果最后改得到的字符串时符合要求的字符串，统计改变的次数，否则返回0. 和图的遍历类似，只是每次要遍历26个字符，用BFS最合适。

参考链接：http://www.cnblogs.com/grandyang/p/4539768.html（强烈推荐这个大佬）

C++代码：

class Solution {
public:
    int ladderLength(string beginWord, string endWord, vector<string>& wordList) {
        unordered_set<string> wordset(wordList.begin(),wordList.end());  //建立一个字典。
        if(!wordset.count(endWord)) return 0;  //如果endWord不在字典上，那无论改多少次，都不能得到这个字符串。
        queue<string> q;
        q.push(beginWord);
        int res = 0;
        while(!q.empty()){
            for(int k = q.size(); k > 0; k--){
                string word = q.front(); q.pop();
                if(word == endWord) return res + 1;
                for(int i = 0; i < word.size(); i++){
                    string newword = word;
                    for(char ch = 'a'; ch <= 'z';ch++){
                        newword[i] = ch;
                        if(wordset.count(newword) && newword!=word){
                            q.push(newword);
                            wordset.erase(newword);  //由于转变后得到的字符串不能与前面的相同，所以得在字典上删除相应的字符串。
                        } 
                    }
                }
            }
            ++res;
        }
        return 0;
    }
};