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。。

class Solution {
public:
    vector<int> findSubstring(string S, vector<string> &L) {
        // Start typing your C/C++ solution below
        // DO NOT write int main() function
        int len = L[0].size();
        int size = L.size();
        map<string, int> T;
        map<string, int> V;
        for (int i = 0; i < L.size(); i++) T[L[i]]++;
        vector<int> ret;
        for (int i = 0; i < S.size()-size*len+1; i++) {
            V.clear();
            int j = 0;
            for (; j < size; j++) {
                string tmp = S.substr(i+j*len, len);
                if (T.find(tmp) != T.end()) V[tmp]++;
                else break;
                if (V[tmp] > T[tmp]) break;
            }
            if (j == size) ret.push_back(i);
        }
        return ret;
    }
};

 C#

public class Solution {
    public List<int> FindSubstring(string s, string[] words) {
        int len = words[0].Length;
        int size = words.Length;
        Dictionary<string, int> T = new Dictionary<string, int>();
        Dictionary<string, int> V = new Dictionary<string, int>();
        for (int i = 0; i < size; i++) {
            if (T.ContainsKey(words[i])) T[words[i]]++;
            else T.Add(words[i], 1);
        }
        List<int> ans = new List<int>();
        for (int i = 0; i < s.Length-size*len+1; i++) {
            V.Clear();
            int j = 0;
            for (; j < size; j++) {
                string tmp = s.Substring(i+j*len, len);
                if (T.ContainsKey(tmp)) {
                    if (V.ContainsKey(tmp)) V[tmp]++;
                    else V.Add(tmp, 1);
                }
                else break;
                if (V[tmp] > T[tmp]) break;
            }
            if (j == size) ans.Add(i);
        }
        return ans;
    }
}