[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.

class Solution {
public:
vector<vector<string> > findLadders(string start, string end, unordered_set<string> &dict)
{
result_.clear();
unordered_map<string, vector<string>> prevMap;
for(auto iter = dict.begin(); iter != dict.end(); ++iter)
prevMap[*iter] = vector<string>();
vector<unordered_set<string>> candidates(2);
int current = 0;
int previous = 1;
candidates[current].insert(start);
while(true)
{
current = !current;
previous = !previous;
for (auto iter = candidates[previous].begin(); iter != candidates[previous].end(); ++iter)
dict.erase(*iter);
candidates[current].clear();

for(auto iter = candidates[previous].begin(); iter != candidates[previous].end(); ++iter)
{
for(size_t pos = 0; pos < iter->size(); ++pos)
{
string word = *iter;
for(int i = 'a'; i <= 'z'; ++i)
{
if(word[pos] == i)continue;
word[pos] = i;
if(dict.count(word) > 0)
{
prevMap[word].push_back(*iter);
candidates[current].insert(word);
}
}
}
}
if (candidates[current].size() == 0)
return result_;
if (candidates[current].count(end)) break;
}
vector<string> path;
GeneratePath(prevMap, path, end);
return result_;
}

private:
void GeneratePath(unordered_map<string, vector<string>> &prevMap, vector<string>& path, const string& word)
{
if (prevMap[word].size() == 0)
{
path.push_back(word);
vector<string> curPath = path;
reverse(curPath.begin(), curPath.end());
result_.push_back(curPath);
path.pop_back();
return;
}
path.push_back(word);
for (auto iter = prevMap[word].begin(); iter != prevMap[word].end(); ++iter)
GeneratePath(prevMap, path, *iter);
path.pop_back();
}
vector<vector<string>> result_;
};