算法描述：

Given two words (beginWord and endWord), and a dictionary's word list, find all shortest transformation sequence(s) 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 an empty list 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:[  ["hit","hot","dot","dog","cog"],  ["hit","hot","lot","log","cog"]]

Example 2:

Input:beginWord = "hit"endWord = "cog"wordList = ["hot","dot","dog","lot","log"]Output: []Explanation: The endWord "cog" is not in wordList, therefore no possible transformation. 解题思路：这道题比较难。深搜加广搜，广搜用于判断是否有通路，并构建连通图。深搜用于获取所有结果。

vector
     
      
       > findLadders(string beginWord, string endWord, vector
       
        & wordList) {        unordered_set
        
          dict(wordList.begin(),wordList.end());        dict.erase(beginWord);        vector
         
          
           > results; queue
           
             que; que.push(beginWord); unordered_map
            
              levels; unordered_map
             
              
               > graph; int level = 0; bool isFound = true; while(!que.empty()){ level++; int levelSize = que.size(); for(int i=0; i< levelSize; i++){ string curWord = que.front(); que.pop(); string copyWord = curWord; for(int i=0; i < curWord.size(); i++){ char t = copyWord[i]; for(char c = 'a'; c <= 'z'; c++){ copyWord[i]=c; if(copyWord == curWord) continue; if(dict.find(copyWord)!=dict.end()){ levels[copyWord]=level+1; graph[curWord].push_back(copyWord); dict.erase(copyWord); if(copyWord==curWord){ isFound = true; }else{ que.push(copyWord); } }else{ auto it = levels.find(copyWord); if(it!=levels.end() && it->second == level+1){ graph[curWord].push_back(copyWord); } } copyWord[i]=t; } } } } if(isFound){ vector
               
                 temp({beginWord}); backtracking(results,temp,graph,beginWord,endWord); } return results; } void backtracking(vector
                
                 
                  >& results,vector
                  
                   & temp,unordered_map
                   
                    > &graph,string beginWord,string endWord){ if(beginWord == endWord){ results.push_back(temp); return; } if(graph.find(beginWord)!=graph.end()){ for(auto curWord:graph[beginWord]){ temp.push_back(curWord); backtracking(results, temp, graph, curWord, endWord); temp.pop_back(); } } }