Word Ladder II

Given two words (start and end), and a dictionary, find all shortest transformation sequence(s) fromstart 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>> res;    void dfsPath(unordered_map<string, unordered_set<string>> &path, vector<string> &temp, const string &key)    {        if(path[key].size() == 0)        {            temp.push_back(key);            vector<string> tempPath = temp;            reverse(tempPath.begin(), tempPath.end());            res.push_back(tempPath);            temp.pop_back();            return ;        }        temp.push_back(key);        for(unordered_set<string>::iterator iter = path[key].begin(); iter != path[key].end(); ++iter)        {            dfsPath(path, temp, *iter);        }        temp.pop_back();    }    vector<vector<string>> findLadders(string start, string end, unordered_set<string> &dict) {        res.clear();        unordered_set<string> current;        unordered_set<string> next;        unordered_map<string, unordered_set<string>> path;        if(dict.count(start) > 0) dict.erase(start);        current.insert(start);        while(current.count(end) == 0 && !dict.empty())        {            for(unordered_set<string>::iterator iter = current.begin(); iter != current.end(); ++iter)            {                string str = *iter;                for(int i = 0; i < str.length(); ++i)                {                    for(char c = 'a'; c <= 'z'; ++c)                    {                        string tmp = str;                        if(tmp[i] == c) continue;                        tmp[i] = c;                        if(dict.count(tmp) > 0)                        {                            path[tmp].insert(str);                            next.insert(tmp);                        }                    }                }            }            if(next.empty()) break;            for(unordered_set<string>::iterator iter = next.begin(); iter != next.end(); ++iter)            {                dict.erase(*iter);            }            current = next;            next.clear();        }        vector<string> temp;        if(current.count(end) > 0) dfsPath(path, temp, end);        return res;    }};