LeetCode: Word Ladder II

看了别人的答案，这题的关键在于数据结构的选择，一开始选择的multimap肯定MLE

 1 class Solution {

 2 public:

 3     void getPath(vector<vector<string> > &ansSet, vector<string> &ans, int now, vector<vector<int> > &pa, vector<string> &que)

 4     {

 5         if (now == -1) 

 6         {

 7             ansSet.push_back(vector<string>(0));

 8             for (int i = ans.size() - 1; i >= 0; i--)

 9                 ansSet.back().push_back(ans[i]);

10             return;

11         }

12         for (int i = 0; i < pa[now].size(); i++)

13         {

14             ans.push_back(que[now]);

15             getPath(ansSet, ans, pa[now][i], pa, que);

16             ans.pop_back();

17         }

18         

19     }

20     vector<vector<string>> findLadders(string start, string end, unordered_set<string> &dict) {

21         // Start typing your C/C++ solution below

22         // DO NOT write int main() function

23         vector<string> que;

24         vector<int> step;

25         vector<vector<int> > pa;

26         unordered_map<string, int> hash;

27         step.push_back(1);

28         que.push_back(start);

29         pa.push_back(vector<int>(0));

30         pa.back().push_back(-1);

31         int head = 0;

32         int bestStep = -1;

33         vector<vector<string> > ans;

34         while (head < que.size())

35         {

36             string now = que[head];

37             int nowStep = step[head];

38             if (bestStep > -1 && nowStep == bestStep) break;

39             for (int i = 0; i < now.size(); i++)

40                 for (char ch = 'a'; ch <= 'z'; ch++)

41                 {

42                     string next = now;

43                     if (next[i] == ch) continue;

44                     next[i] = ch;

45                     if (next == end)

46                     {

47                         bestStep = nowStep + 1;

48                         vector<string> single(0);

49                         single.push_back(end);

50                         getPath(ans, single, head, pa, que);

51                     } else if (dict.count(next))

52                     {

53                         if (hash[next] == 0)

54                         {

55                             que.push_back(next);

56                             step.push_back(nowStep + 1);

57                             pa.push_back(vector<int>(0));

58                             pa.back().push_back(head);

59                             hash[next] = pa.size() - 1;

60                         } else if (step[hash[next]] == nowStep + 1)

61                         {

62                             pa[hash[next]].push_back(head);

63                         }

64                     }

65                 }

66             head++;

67         }

68         return ans;

69     }

70 };