[leetcode]Word Ladder II

问题描述：

Given two words (start and end), and a dictionary, find all shortest transformation sequence(s) fromstart 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"]

Return

  [    ["hit","hot","dot","dog","cog"],    ["hit","hot","lot","log","cog"]  ]

Note:

• All words have the same length.
• All words contain only lowercase alphabetic characters.

代码：

    void createResult(map<string,vector<string> >&record,vector<vector<string> >&result,string pre,string start,list<string>myList){  //C++        if(pre == start){            vector<string> vec(myList.begin(),myList.end());            result.push_back(vec);            // for(int i=0; i<myList.size();i++)            //     vec.push_back(myList[i]);        }        vector<string> next = record[pre];        for(int i=0; i<next.size(); i++){            string word = next[i];            list<string> temp(myList);            temp.push_front(word);            createResult(record,result,word,start,temp);        }    }    vector<vector<string>> findLadders(string start, string end, unordered_set<string> &dict) {        vector<vector<string> > result;        int size = dict.size();        if(start.size() ==0 || end.size() ==0 || size==0 || start==end)            return result;        map<string,int> myMap;        map<string,vector<string> > record;        queue<string> myQueue;        myQueue.push(start);        myMap[start] = 1;        string temp;                while(!myQueue.empty()){            temp = myQueue.front();            myQueue.pop();                        for(int i =0; i<temp.size(); i++){                string newword = temp;                for(int j=0; j<26; j++){                    newword[i] = 'a'+j;                    if(dict.count(newword)!=0){                        if(myMap.count(newword)==0)                        {                            myMap[newword] = myMap[temp] +1;                            myQueue.push(newword);                            vector<string> vec;                            vec.push_back(temp);                            record.insert(make_pair(newword,vec));                        }                        else if(myMap[newword] == myMap[temp]+1){                            vector<string> vec = record[newword];                            vec.push_back(temp);                            record[newword]=vec;                        }                    }                }            }                    }        if(record.count(end)==0)            return result;                    list<string> temp_list;        temp_list.push_back(end);        createResult(record,result,end,start,temp_list);    }