leetcode 30. Substring with Concatenation of All Words

Input:
  s = "barfoothefoobarman",
  words = ["foo","bar"]
Output: [0,9]

这道题给定的words中的单词长度都是一样的，从连续的s子字符串中找出words中的所有单词，并且words中对应的连续的s是不能重叠的，words的顺序可以打乱；

1、从s中找出所有words中存在的单词

2、在长度为words.size()*words[0].length()的s中，是否存在words中所有的单词，并且不重叠；

这里给的关键信息是words中的单词都是一样长的，所以可以通过定长的方式进行匹配；

    vector<int> findSubstring(string s, vector<string>& words) {
        vector<int> result;
        if (words.size() == 0)
            return result;
        else if (words[0].length() == 0)
            return result;
        else if (s.length() < words.size()*words[0].length())
            return result;
        
        map<string, int> word_count;
        for (auto iter = words.begin(); iter != words.end(); ++iter) {
            auto index = word_count.find(*iter);
            if (index != word_count.end())
                (index->second)++;
            else
                word_count[*iter] = 1;
        }
        for (int i = 0; i < words[0].length(); ++i) {
            map<string, int> s_count;
            int count = 0;
            for (int j = i; j < s.length() - words[0].length() + 1; j += words[0].length()) {
                auto index = word_count.find(s.substr(j, words[0].length()));
                if (index == word_count.end()) {
                    s_count.clear();
                    count = 0;
                    continue;
                }
                ++count;
                auto s_index = s_count.find(index->first);
                if (s_index != s_count.end())
                    (s_index->second)++;
                else
                    s_count[index->first] = 1;
                if (count == words.size() && s_count == word_count) {
                    result.push_back(int(j - words[0].length() * (words.size()-1)));
                }

                if (count == words.size()) {
                    s_index = s_count.find(s.substr(j - words[0].length() * (words.size()-1), words[0].length()));
                    (s_index->second)--;
                    if (s_index->second == 0)
                        s_count.erase(s_index);
                    --count;
                }
            }
        }
        
        return result;
    }