126. Word Ladder II

思路：BFS+DFS
首先和Word Ladder一样，先BFS构建一个梯子，例如，用string_map将[“hot”,“dot”,“dog”,“lot”,“log”,“cog”]映射为1，2，3，2，3，4
然后用DFS从endWord开始往回构建vector对象，像下楼梯一样一步一步走回来4->3->2->1

看了题解，不然我连第一题都不会的哈哈

class Solution {
public:
    vector<vector<string>>res;
    void dfs(vector<string>temp_res,string str,string beginWord,unordered_map<string,int>string_map){
        int next_position=string_map[str]-1;
        for(int i=0;i<beginWord.length();++i){
            string current=str;
            for(char ch='a';ch<='z';++ch){
                current[i]=ch;
                if(current==str)continue;
                if(string_map[current]==next_position){
                    temp_res.push_back(current);
                    if(current==beginWord){
                        reverse(temp_res.begin(),temp_res.end());
                        res.push_back(temp_res);
                        return;
                    }
                    dfs(temp_res,current,beginWord,string_map);
                    temp_res.pop_back();
                }
            }
        }
    }
    vector<vector<string>> findLadders(string beginWord, string endWord, vector<string>& wordList) {
        unordered_set<string>string_set(wordList.begin(),wordList.end());
        if(!string_set.count(endWord))return res;
        unordered_map<string,int>string_map{{beginWord,1}};
        queue<string>q;
        q.push(beginWord);
        //bfs
        int end_position=INT_MAX;
        while(!q.empty()){
            string front_string=q.front();
            if(string_map[front_string]>=end_position)break;
            q.pop();
            for(int i=0;i<beginWord.length();++i){
                string current=front_string;
                for(char ch='a';ch<='z';++ch){
                    current[i]=ch;

                    if(current==front_string)continue;
                    if(current==endWord)end_position=string_map[current]+1;
                    if(string_set.count(current)&&string_map[current]==0){
                        q.push(current);
                        string_map[current]=string_map[front_string]+1;
                    }

                }
            }
        }
        //dfs
        vector<string>temp_res;
        temp_res.push_back(endWord);
        dfs(temp_res,endWord,beginWord,string_map);
        return res;
    }
};