[LeetCode] 把一个单词变形成为另一个单词，每次变形只改变一个单词 word ladder

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给定两个相等长度的单词，写一个算法，把一个单词变形成为另一个单词，每次变形只改变一个单词。此过程中得到的每个单词必须是字典里存在的单词。

EXAMPLE
Input: DAMP, LIKE
Output: DAMP -> LAMP -> LIMP -> LIME -> LIKE

思路：

这其实就是广度优先遍历。遍历过程中，我们需要记录路径，在找到目标单词后，根据记录的路径打印出变形过程。

    int ladderLength(string start, string end, unordered_set<string> &dict)     {// The driver function        map<string, vector<string>> graph; // The graph                // Build the graph according to the method given in:        //     http://yucoding.blogspot.com/2013/08/leetcode-question-127-word-ladder.html        for(unordered_set<string>::iterator it = dict.begin(); it!=dict.end(); it++)        {       graph[*it] = findDict(*it, dict); // find out the neighbors of *it        }                // Do breadth first traversal        return BFT(start, end, graph);            }        int BFT(string start, string end, map<string, vector<string>>& graph)    {        deque<pair<string,int>> queue; // The queue for holding the nodes in the graph        unordered_set<string> visited; // Mark those nodes that have been visited                queue.push_back( pair<string, int>(start,1) ); // push back the string and its level                while(queue.size()>0)        {            pair<string,int> front = queue.front();            queue.pop_front();                        if(front.first==end)                return front.second;                            vector<string> vec = graph[front.first];                        for(int i=0; i<vec.size(); i++)            {                if(visited.find(vec[i])==visited.end())                {                    visited.insert(vec[i]);                    queue.push_back(pair<string,int>(vec[i], front.second+1) );                }            }        }                return 0;    }        vector<string> findDict(string str, unordered_set<string> &dict){    // Find out the neighbors of str in dict        vector<string> res;        int sz = str.size();        string s = str;        for (int i=0;i<sz;i++){            s = str;            for (char j = 'a'; j<='z'; j++){                s[i]=j;                if (dict.find(s)!=dict.end()){                    res.push_back(s);                }            }                     }        return res;    }