LeetCode-30-Substring with Concatenation of All Words

算法描述：

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

Example 1:

Input:
  s = "barfoothefoobarman",
  words = ["foo","bar"]
Output: [0,9]
Explanation: Substrings starting at index 0 and 9 are "barfoor" and "foobar" respectively.
The output order does not matter, returning [9,0] is fine too.

Example 2:

Input:
  s = "wordgoodgoodgoodbestword",
  words = ["word","good","best","word"]
Output: []

解题思路：用两个map分别记录字典单词数量和使用的单词数量。当单词的使用量大于字典数量，或者连续截取单词不在字典中，则返回失败。

vector<int> findSubstring(string s, vector<string>& words) {
        vector<int> results;
        if(s.size()==0 || words.size()==0) return results;
        unordered_map<string, int> map;
        for(auto word:words){
            map[word]++;
        }
        int m = words.size();
        int n= words[0].size();
        for(int i=0; i < s.size() - m*n +1; i++){
            unordered_map<string, int> seen;
            int j =0;
            while(j < m){
                string temp = s.substr(i+j*n,n);
                if(map.find(temp)!=map.end()){
                    seen[temp]++;
                    if(seen[temp] >map[temp]) break;
                } else
                    break;
                j++;
            }
            if(j==m) results.push_back(i);
        }
        return results;
    }