LeetCode: Substring with Concatenation of All Words

自己想的大体思路还是对的，不过可以改进的很多，最后网上找了个勉勉强强过large的答案，用map实在太容易超时了。。

 1 class Solution {

 2 public:

 3     vector<int> findSubstring(string S, vector<string> &L) {

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

 5         // DO NOT write int main() function

 6         int len = L[0].size();

 7         int size = L.size();

 8         map<string, int> T;

 9         map<string, int> V;

10         for (int i = 0; i < L.size(); i++) T[L[i]]++;

11         vector<int> ret;

12         for (int i = 0; i < S.size()-size*len+1; i++) {

13             V.clear();

14             int j = 0;

15             for (; j < size; j++) {

16                 string tmp = S.substr(i+j*len, len);

17                 if (T.find(tmp) != T.end()) V[tmp]++;

18                 else break;

19                 if (V[tmp] > T[tmp]) break;

20             }

21             if (j == size) ret.push_back(i);

22         }

23         return ret;

24     }

25 };

 C#

 1 public class Solution {

 2     public List<int> FindSubstring(string s, string[] words) {

 3         int len = words[0].Length;

 4         int size = words.Length;

 5         Dictionary<string, int> T = new Dictionary<string, int>();

 6         Dictionary<string, int> V = new Dictionary<string, int>();

 7         for (int i = 0; i < size; i++) {

 8             if (T.ContainsKey(words[i])) T[words[i]]++;

 9             else T.Add(words[i], 1);

10         }

11         List<int> ans = new List<int>();

12         for (int i = 0; i < s.Length-size*len+1; i++) {

13             V.Clear();

14             int j = 0;

15             for (; j < size; j++) {

16                 string tmp = s.Substring(i+j*len, len);

17                 if (T.ContainsKey(tmp)) {

18                     if (V.ContainsKey(tmp)) V[tmp]++;

19                     else V.Add(tmp, 1);

20                 }

21                 else break;

22                 if (V[tmp] > T[tmp]) break;

23             }

24             if (j == size) ans.Add(i);

25         }

26         return ans;

27     }

28 }

View Code