Word Ladder II

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.

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;
}
};