ssss

class Solution {
public:
	vector<int> findSubstring(string S, vector<string> &L) {
		unordered_map<string, int>wordTimes;//L中单词出现的次数
		for (int i = 0; i < L.size(); i++){
			if (wordTimes.count(L[i]) == 0)
				wordTimes.insert(make_pair(L[i], 1));
			else wordTimes[L[i]]++;
		}
		int wordLen = L[0].size();

		vector<int> res;
		for (int i = 0; i < wordLen; i++)
		{//为了不遗漏从s的每一个位置开始的子串，第一层循环为单词的长度
			unordered_map<string, int>wordTimes2;//当前窗口中单词出现的次数
			int winStart = i, cnt = 0;//winStart为窗口起始位置,cnt为当前窗口中的单词数目
			for (int winEnd = i; winEnd <= (int)S.size() - wordLen; winEnd += wordLen)
			{//窗口为[winStart,winEnd)
				string word = S.substr(winEnd, wordLen);
				if (wordTimes.find(word) != wordTimes.end())
				{
					if (wordTimes2.find(word) == wordTimes2.end())
						wordTimes2[word] = 1;
					else wordTimes2[word]++;

					if (wordTimes2[word] <= wordTimes[word])
						cnt++;
					else
					{//当前的单词在L中，但是它已经在窗口中出现了相应的次数，不应该加入窗口
						//此时，应该把窗口起始位置想左移动到，该单词第一次出现的位置的下一个单词位置
						for (int k = winStart;; k += wordLen)
						{
							string tmpstr = S.substr(k, wordLen);
							wordTimes2[tmpstr]--;
							if (tmpstr == word)
							{
								winStart = k + wordLen;
								break;
							}
							cnt--;
						}
					}

					if (cnt == L.size())

						res.push_back(winStart);
				}
				else
				{//发现不在L中的单词
					winStart = winEnd + wordLen;
					wordTimes2.clear();
					cnt = 0;
				}
			}
		}
		return res;
	}
};