本題有幾個注意點:
1. 回溯找路徑時,根據路徑的最大長度控制回溯深度
2. BFS時,在找到end單詞後,給當前層做標記find=true,遍歷完當前層後結束。不需要遍歷下一層了。
3. 可以將字典中的單詞刪除,替代visited的set,這樣優化以後時間從1700ms+降到800ms+
代碼如下:
class Solution { public: vector> findLadders(string start, string end, unordered_set &dict) { set queue[2]; queue[0].insert(start); vector > res; bool find = false; int length = 1; bool cur = false; map > mapping; //bfs while (queue[cur].size() && !find) { length++; for (set ::iterator i = queue[cur].begin(); i != queue[cur].end(); i++)//delete from dictionary dict.erase(*i); for (set ::iterator i = queue[cur].begin(); i != queue[cur].end(); i++) { for (int l = 0; l < (*i).size(); l++) { string word = *i; for (char c = 'a'; c <= 'z'; c++) { word[l] = c; if (dict.find(word) != dict.end()) { if (mapping.find(word) == mapping.end()) mapping[word] = set (); mapping[word].insert(*i); if (word == end) find = true; else queue[!cur].insert(word); } } } } queue[cur].clear(); cur = !cur; } if (find) { vector temp; temp.push_back(end); getRes(mapping, res, temp, start, length); } return res; } void getRes(map > & mapping, vector > & res, vector temp, string start, int length) { if (temp[0] == start) { res.push_back(temp); return; } if (length == 1) return;//recursion depth string word = temp[0]; temp.insert(temp.begin(), ""); for (set ::iterator j = mapping[word].begin(); j != mapping[word].end(); j++) { temp[0] = *j; getRes(mapping, res, temp, start, length - 1); } } };