This is a tough problem! I read some solutions from the web. This link provides a one-end BFS solution. The code is reorganized as follows.
1 class Solution { 2 public: 3 vector<vector<string>> findLadders(string start, string end, unordered_set<string> &dict) { 4 dict.insert(end); 5 unordered_set<string> current; 6 unordered_set<string> next; 7 current.insert(start); 8 ladder.push_back(end); 9 while (true) { 10 for (string cur : current) 11 dict.erase(cur); 12 for (string cur : current) 13 findChildren(cur, next, dict); 14 if (next.empty()) return ladders; 15 if (next.find(end) != next.end()) { 16 genLadders(start, end); 17 return ladders; 18 } 19 current.clear(); 20 swap(current, next); 21 } 22 } 23 private: 24 unordered_map<string, vector<string> > ancestors; 25 vector<string> ladder; 26 vector<vector<string> > ladders; 27 28 void findChildren(string& word, unordered_set<string>& next, unordered_set<string>& dict) { 29 int len = word.length(); 30 string ancestor = word; 31 for (int p = 0; p < len; p++) { 32 char letter = word[p]; 33 for (int i = 0; i < 26; i++) { 34 word[p] = 'a' + i; 35 if (dict.find(word) != dict.end()) { 36 next.insert(word); 37 ancestors[word].push_back(ancestor); 38 } 39 } 40 word[p] = letter; 41 } 42 } 43 44 void genLadders(string& start, string& end) { 45 if (start == end) { 46 reverse(ladder.begin(), ladder.end()); 47 ladders.push_back(ladder); 48 reverse(ladder.begin(), ladder.end()); 49 return; 50 } 51 for (string ancestor : ancestors[end]) { 52 ladder.push_back(ancestor); 53 genLadders(start, ancestor); 54 ladder.pop_back(); 55 } 56 } 57 };
This link shares a two-end BFS solution, which is much faster! I have also rewritten that code below.
1 class Solution { 2 public: 3 vector<vector<string>> findLadders(string start, string end, unordered_set<string> &dict) { 4 vector<string> ladder; 5 vector<vector<string> > ladders; 6 ladder.push_back(start); 7 unordered_set<string> startWords; 8 unordered_set<string> endWords; 9 startWords.insert(start); 10 endWords.insert(end); 11 unordered_map<string, vector<string> > children; 12 bool flip = true; 13 if (searchLadders(startWords, endWords, children, dict, ladder, ladders, flip)) 14 genLadders(start, end, children, ladder, ladders); 15 return ladders; 16 } 17 private: 18 bool searchLadders(unordered_set<string>& startWords, unordered_set<string>& endWords, 19 unordered_map<string, vector<string> >& children, unordered_set<string>& dict, 20 vector<string>& ladder, vector<vector<string> >& ladders, bool& flip) { 21 flip = !flip; 22 if (startWords.empty()) return false; 23 if (startWords.size() > endWords.size()) 24 return searchLadders(endWords, startWords, children, dict, ladder, ladders, flip); 25 for (string word : startWords) dict.erase(word); 26 for (string word : endWords) dict.erase(word); 27 unordered_set<string> intermediate; 28 bool done = false; 29 for (string word : startWords) { 30 string temp = word; 31 int len = word.length(); 32 for (int p = 0; p < len; p++) { 33 char letter = word[p]; 34 for (int i = 0; i < 26; i++) { 35 word[p] = 'a' + i; 36 if (endWords.find(word) != endWords.end()) { 37 done = true; 38 flip ? children[word].push_back(temp) : children[temp].push_back(word); 39 } 40 else if (!done && dict.find(word) != dict.end()) { 41 intermediate.insert(word); 42 flip ? children[word].push_back(temp) : children[temp].push_back(word); 43 } 44 } 45 word[p] = letter; 46 } 47 } 48 return done || searchLadders(endWords, intermediate, children, dict, ladder, ladders, flip); 49 } 50 void genLadders(string& start, string& end, unordered_map<string, vector<string> >& children, 51 vector<string>& ladder, vector<vector<string> >& ladders) { 52 if (start == end) { 53 ladders.push_back(ladder); 54 return; 55 } 56 for (string child : children[start]) { 57 ladder.push_back(child); 58 genLadders(child, end, children, ladder, ladders); 59 ladder.pop_back(); 60 } 61 } 62 };
