LeetCode-Word LadderII

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.

和Word Ladder不同的地方在于可能多个父节点公用一个子节点。将父节点都记录下来即可。

class Solution {
public:
		struct info{
			info(){}
			info(int level,int index){
				m_level=level;
				m_index=index;
			}
			int m_level;
			int m_index;
		};
		void Path(vector<vector<string> >* ret,vector<string>* path,const vector<vector<int> >&father,const vector<string>& record,int index,int count){
			(*path)[count]=record[index];
			if(count==0){
				ret->push_back(*path);
			}
			else{
				for(int i=0;i<father[index].size();i++){
					Path(ret,path,father,record,father[index][i],count-1);
				}
			}
		}
		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
			map<string,info> m;
			int qhead=0;
			vector<string> record;
			vector<int> tails;
			vector<vector<int> >father;
			int index=0;
			m[start]=info(0,0);
			record.push_back(start);
			father.resize(father.size()+1);
			int min_v=2147483647;
			
			while(qhead<record.size()!=0){
				int currentIndex=index;
				string s=record[qhead];
				for(int i=0;i<s.length();i++){
					char c=s[i];
					for(int j='a';j<'z';j++){
						if(j!=c){
							s[i]=j;
							if(s==end){
								int level=m[record[qhead]].m_level+1;
								if(level<min_v){
									min_v=level;
									tails.clear();
									tails.push_back(qhead);
								}
								else if(level==min_v){
									tails.push_back(qhead);
								}
							}
							else if(dict.find(s)!=dict.end()){
								if(m.find(s)==m.end()){
									index++;
									m[s]=info(m[record[qhead]].m_level+1,index);
									father.resize(father.size()+1);
									father[index].push_back(qhead);
									record.push_back(s);
								}
								else{
									info sinfo=m[s];
									info tinfo=m[record[qhead]];
									if(sinfo.m_level==tinfo.m_level+1){
										father[sinfo.m_index].push_back(qhead);
									}
								}
							}
						}
						s[i]=c;
					}
				}
				qhead++;
			}
			if(min_v==2147483647){
				return vector<vector<string> >();
			
			}

			vector<vector<string> >ret;
			vector<string>path;

			path.resize(min_v+1);
			path[min_v]=end;
			for(int i=0;i<tails.size();i++){
				Path(&ret,&path,father,record,tails[i],min_v-1);
			}
			return ret;
	}
};