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).

思路：题目的意思是找到包含L中所有单词的起始位置，但是不能多，也不能少，要刚好包含L中的全部单词，而且相邻的两个开始位置还可以重叠，比如：S="aaa",L=[‘a‘,‘a‘]，结果是[0,1]。具体方法和最短摘要有点类似，用need数组保存需要的所有单词，hasFind保存已经找到的单词，如果hasFind大于need，则该位置不满足，只有当两个数组完全相同时才满足。

class Solution {
public:
    vector<int> findSubstring(string S, vector<string> &L)
    {
    	int totalWord = L.size(),wordLen = L[0].size(),i,j;
    	map<string,int> needWord;	
    	vector<int> res;
    	for (i = 0;i < totalWord;i++)needWord[L[i]]++;//统计需要的单词
    	for(i = 0;i + totalWord*wordLen <= S.size();i++)
    	{
    		map<string,int> hasFindWord;
    		for (j =0;j < totalWord;j++)//对于每一个单词
    		{
    			string sub = S.substr(i+j*wordLen,wordLen);//提取每一个可能的单词
    			map<string,int>::iterator iter = needWord.find(sub);
    			if(iter == needWord.end())break;
    			hasFindWord[sub]++;
    			if(hasFindWord[sub] > needWord[sub])break;
    		}
    		if(j == totalWord)	res.push_back(i);
    	}
    	return res;
    }

};

leetcode 之 Substring with Concatenation of All Words