Word Ladder II

-----QUESTION-----

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

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.

-----SOLUTION-----

class Solution {public:    vector<vector<string>> findLadders(string start, string end,unordered_set< string > &dict){        result.clear();        unordered_map<string, vector<string>> adj;        unordered_set<string> set_to_push; //可能存在两个word的pre相同，如果用queue会有重复，所以使用set        queue<string> que_to_pop;        que_to_pop.push(end);        string current, tmp;        bool stopFlag = false;        vector word_to_delete;        word_to_delete.push_back(start);        word_to_delete.push_back(end);                //find the shortest path from start to end        while(!que_to_pop.empty()){            current = que_to_pop.front();            que_to_pop.pop();            for(int i = 0; i < current.size(); i++){                tmp = current;                for(char c = 'a'; c <= 'z'; c++){                    if(current[i]==c) continue;                    tmp[i] = c;                    if(tmp == start){                        stopFlag = true;                        adj[tmp].push_back(current);                    }                    else if(!stopFlag && dict.find(tmp)!=dict.end()) {                        set_to_push.insert(tmp);                        adj[tmp].push_back(current);                        word_to_delete.push_back(tmp);                    }                }            }            if(que_to_pop.empty()) { // end a level                if(stopFlag) break;                for(unordered_set< string >::iterator it = set_to_push.begin(); it!=set_to_push.end(); it++){                    que_to_pop.push(*it);                }                set_to_push.clear();                for(int i = 0; i < word_to_delete.size(); i++) dict.erase(word_to_delete[i]);                word_to_delete.clear();            }        }                //push in the result        vector< string > item;        pushResult(start, end, adj, item);        return result;    }        void pushResult(string pre, string end, unordered_map<string, vector<string>>& adj, vector<string>& item){        if(pre == end){            item.push_back(end);            result.push_back(item);            item.pop_back();            return;        }        for(int i = 0 ; i < adj[pre].size(); i++){            item.push_back(pre);            pushResult(adj[pre][i], end, adj,item);            item.pop_back();        }    }    private:    vector<vector<string>> result;};

BFS：Memory limitexceeded

class Solution {public:    vector<vector<string>> findLadders(string start, string end, unordered_set<string>&dict) {        int wordLen = start.length();        string current;        queue<string> queue_to_push; //prepared for next loop        queue<string> queue_to_pop; //string to deal in this loop        vectorvStr;        vector< vector< string >> result;        bool endFlag = false;              vStr.push_back(start);        queue_to_pop.push(vStr);        while(!queue_to_pop.empty())        {            vStr = queue_to_pop.front();            queue_to_pop.pop();            current = *(vStr.end()-1);                     for (int i = 0; i < wordLen; i++){                int j = 'a';                for (; j <= 'z'; j++)                  {                      if (j == current[i]) //skip the current character                          continue;                      char temp = current[i];                      current[i] = j;                      if(current==end)                    {                        vStr.push_back(current);                        result.push_back(vStr);                        endFlag = true;                        break;                    }                    if (!endFlag && dict.count(current) > 0) //exist such a word in the dict                      {                          vStr.push_back(current);                        queue_to_push.push(vStr);                        vStr.pop_back();                    }                      current[i] = temp;  //restore the string                }                if(j <= 'z') break;            }                            if(queue_to_pop.empty())            {                if(endFlag) break;                swap(queue_to_push, queue_to_pop);            }        }        return result;    }};