leetcode 126: Word Ladder ll (uncompleted)

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Word Ladder IIFeb 11

Given two words (start and end), and a dictionary, find all shortest transformation sequence(s) from start to end, such that:

Only one letter can be changed at a time
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.

class Solution {public:    vector<vector<string> > findLadders(string start, string end, unordered_set<string> &dict) {        // Start typing your C/C++ solution below        // DO NOT write int main() function                vector<vector<string> > res;        if(start.size()!=end.size()) return res;       // if(start == end) return ;                unordered_set<string> unique;        unique.insert(start);                queue<string> que;        unordered_map<string, string> path;                que.push(start);        path[start] = "";        int q1=1;        int q2=0;                while(q1>0) {            string s = que.front();            que.pop();            --q1;            string pre = "";            for(int i=0; i<s.size(); i++) {                string temp = s;                for(char c='a'; c<='z'; c++) {                    temp[i] = c;                    if(dict.find(temp)!=dict.end() && unique.find(temp)==dict.end()) {                        if(temp == end) {                            if(pre!="") {                                unique.erase(pre);                            }                            pre = temp;                            vector<string> elems;                            elems.push_back(end);                            string item = s;                            while(item!="") {                                elems.insert(elems.begin(), item);                                item = path[item];                            }                            res.push_back(elems);                        } else {                            unique.insert(temp);                            que.push(temp);                            path[temp] = s;                            ++q2;                        }                    }                }            }                        if(q1==0) {                if(!res.empty()) return res;                q1 = q2;                q2 = 0;            }        }        return res;    }};

class Solution {public:     class TreeNode{        public:            string val;            TreeNode * parent;            TreeNode(string str, TreeNode* p=NULL):val(str), parent(p){}          };    bool findnode(string temp,  TreeNode* p) {        while(p!=NULL) {            if( temp == p->val) return false;            p = p->parent;        }        return true;    }    vector<vector<string> > findLadders(string start, string end, unordered_set<string> &dict) {        // Start typing your C/C++ solution below        // DO NOT write int main() function        vector<TreeNode*> clean;        vector<vector<string> > res;        if(start.size()!=end.size()) return res;               // if(start == end) return ;        queue<TreeNode*> que;        TreeNode *head = new TreeNode(start);        clean.push_back(head);        TreeNode *p = head;        que.push(p);        int q1=1;        int q2=0;                while(q1>0) {            TreeNode * p = que.front();            string s = p->val;            que.pop();            --q1;                        for(int i=0; i<s.size(); i++) {                string temp = s;                for(char c='a'; c<='z'; c++) {                    temp[i] = c;                    if(dict.find(temp)!=dict.end() && findnode(temp, p) ) {                        if(temp == end) {                            vector<string> elems;                            elems.push_back(end);                            while(p!=NULL) {                                elems.insert(elems.begin(), p->val);                                p = p->parent;                            }                            res.push_back(elems);                        } else {                            TreeNode* tn = new TreeNode(temp,p);                            clean.push_back(tn);                            que.push(tn);                            ++q2;                        }                    }                }            }                        if(q1==0) {                if(!res.empty()) return res;                q1 = q2;                q2 = 0;            }        }                for(int i=0; i<clean.size(); i++) {            delete clean[i];            clean[i] = NULL;        }        return res;    }};