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.

思路：

字符串的花儿挺多，word ladder也是CC150 v5上的18.10题，乍一看无从下手，其实就是BFS求单源无权最短路径。

想要大set不超时，关键有几点：

1. 一边加新词进queue，一边从dict里remove掉

2. 直接寻找所有可能的OneEditWords判断是否在dict里面，每次都过一遍dict一个一个判断是否为OneEditWords会超时。我还有一个超时的方法，就是寻找所有可能的OneEditWords再建一个跟dict求交集，后来发现algoritm里的只支持两个有序集合很坑，需要先把unordered_set转化为vector排序，结果不言而喻。

3. 将<string, int > path并到访问队列qwords里去会跑得更快，反正这里不要求输出路径，这个可以做下优化

4. 将getOneEditWords这个函数并到ladderLength里头会跑得更快，不过我觉得可读性会降低

1 int ladderLength(string start, string end, unordered_set
     
       &dict) { 2     unordered_set
      
        ndict(dict);//最好不要修改输入参数 3     queue
       
         qwords;//BFS访问队列 4     unordered_map
        
          path;//记录到start的最短路径，其实可以并入qwords中 5      6     qwords.push(start); 7     path[start] = 1; 8     ndict.erase(start); 9     10     while(!qwords.empty()){11         start = qwords.front();12         qwords.pop();13         int len = path[start];14         for(string s :getOneEditWords(start)){
   //边局部构建map，边处理15             if(ndict.find(s) == ndict.end()) continue;//一个一个判断dict中元素是否为OneEditWords会超时16             if(s == end) return len+1;17             qwords.push(s);18             path[s] = len+1;19             ndict.erase(s);//如果不erase访问过的元素会超时20         }21     }22     return 0;    23 } 24 25 vector
         
           getOneEditWords(string start){26     vector
          
            words;27 for(unsigned int i = 0; i < start.size(); i++){28 string word(start);29 for(char ch = 'a'; ch <= 'z'; ch++){30 if(ch != start[i]){31 word[i] = ch;32 words.push_back(word);33 }34 }35 }36 return words; 37 }