Word Ladder

Given two words (start and end), and a dictionary, find the length of shortest transformation sequence 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"]

As one shortest transformation is "hit" -> "hot" -> "dot" -> "dog" -> "cog",
return its length 5.

Note:

• Return 0 if there is no such transformation sequence.
• All words have the same length.
• All words contain only lowercase alphabetic characters.

struct Node

{

    string word;

    int path;

};



class Solution {

public:

    int ladderLength(string start, string end, unordered_set<string> &dictout) 

    {

        if(start==end) return 1;

        

        int len=start.length();

        dictout.insert(start);

        dictout.insert(end);

        

        vector<Node*> path;

        

        Node* newnode=new Node();

        newnode->word=start;newnode->path=1;

        path.push_back(newnode);

        

        unordered_set<string> dictin;

        dictin.insert(start);

        dictout.erase(dictout.find(start));

        

        int index=0;

        while(index<path.size())

        {

            if(path[index]->word==end) return path[index]->path;

            //generate next

            string news=path[index]->word;

            for(int i=0;i<len;i++)

                for(int k=0;k<26;k++)

                    if(path[index]->word[i]!=k+'a')

                    {

                        news[i]=k+'a';

                        unordered_set<string>::const_iterator findout=dictout.find(news);

                        if(findout!=dictout.end() && dictin.find(news)==dictin.end())

                        {

                            newnode=new Node();

                            newnode->path=path[index]->path+1;

                            newnode->word=news;

                            path.push_back(newnode);

                            

                            dictin.insert(news);

                            dictout.erase(findout);

                        }

                        news[i]=path[index]->word[i];

                    }

            

            delete path[index];

            index++;

        }

        return 0;

    }

};