leetcode 126 —— 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.

思路： 这道题目的难度简直令人发指。参考这里

class Solution {
public:
vector<vector<string>> res;
vector<string> tmpPath;
unordered_map<string, unordered_set<string>> path;
void generatePath(const string& start, string& end)
{
tmpPath.push_back(start);
if (start == end){
vector<string> tmp = tmpPath;
reverse(tmp.begin(), tmp.end());
res.push_back(tmp);
return;
}
for (auto index = path[start].begin(); index != path[start].end(); index++){
generatePath(*index, end);
tmpPath.pop_back();
}
}
vector<vector<string>> findLadders(string start, string end, unordered_set<string> &dict) {
unordered_set<string> cur;
unordered_set<string> nex;
unordered_set<string> unvisited=dict;
if (unvisited.count(start) > 0)
unvisited.erase(start);
cur.insert(start);
while (cur.count(end) == 0 && !unvisited.empty()){
for (auto index = cur.begin(); index != cur.end(); index++){
string word = *index;
for (int i = 0; i < word.size(); i++){
for (int j = 0; j < 26; j++){
string tmp = word;
tmp[i] = 'a' + j;
if (unvisited.count(tmp)>0){
path[tmp].insert(word);
nex.insert(tmp);
}
}
}
}
if (nex.empty()) break;

for (auto index = nex.begin(); index != nex.end(); index++){
unvisited.erase(*index);
}

cur = nex;
nex.clear();
}
if (cur.count(end) > 0){
generatePath( end, start);
}
return res;
}
};