Leetcode: Substring with Concatenation of All Words

You are given a string, S, and a list of words, L, that are all of the same length. Find all starting indices of substring(s) in S that is a concatenation of each word in L exactly once and without any intervening characters.

For example, given:
S: "barfoothefoobarman"
L: ["foo", "bar"]

You should return the indices: [0,9].
(order does not matter).

一开始理解错题意了，以为只要有L中的单词在S中没有干扰词汇的连续出现就要记下来呢，提交发现不对，人家是each word in L exactly once.是要所有L中的词都出现一遍，中间不能有其他的干扰词汇。并且L中的词也会有重复。

错误代码：

vector<int> findSubstring(string S, vector<string> &L) {
        // Note: The Solution object is instantiated only once.
		set<string> dic;
		for(int i = 0; i < L.size(); i++)
			dic.insert(L[i]);
 
		vector<int> res;
		int wordlen = L[0].size();
		set<string>::iterator it;
		int i = 0;
		bool isLastOk = false;
		while(i < S.size())
		{
			string sub = S.substr(i,wordlen);
			it = dic.find(sub);
			if(it != dic.end())
			{
				if(!isLastOk)
				{
					res.push_back(i);
					isLastOk = true;
				}
					i += wordlen;
			}else
			{
				isLastOk = false;
				i++;
			}
		}
		return res;
    }

正确代码如下：

vector<int> findSubstring(string S, vector<string> &L) {
        // Note: The Solution object is instantiated only once.
		map<string,int> words;
		map<string,int> cur;
		int wordNum = L.size();
		for(int i = 0; i < wordNum; i++)
			words[L[i]]++;
		int wordLen = L[0].size();
		vector<int> res;
		//if(S.size() < wordLen*wordNum)return res;
		for(int i = 0; i <= (int)S.size()-wordLen*wordNum; i++)
		{
			cur.clear();
			int j;
			for(j = 0; j < wordNum; j++)
			{
				string word = S.substr(i+j*wordLen, wordLen);
				if(words.find(word) == words.end())
					break;
				cur[word]++;
				if(cur[word]>words[word])
					break;
			}
			if(j == wordNum)
				res.push_back(i);
		}
		return res;
	}