Substring with Concatenation of All Words[leetcode]

思路是用一个map记录词典l中所有单词以及出现次数。

记录单词长度wl,词典中单词总长度ttl。

对于所有可行的开始位置，在O(s.size)时间内查找是否有满足要求的字符串。关键点为找到map[substr]==0，以及不存在的map[substr]

总的时间复杂度为O(s.size * wl)

vector<int> findSubstring(string s, vector<string> &l) {        //init        map<string, int> wordMap;        vector<int> res;        for (int i = 0; i < l.size();i++)        {            if (wordMap.find(l[i]) == wordMap.end()) wordMap[l[i]] = 0;            wordMap[l[i]] ++;        }        int wl = l[0].size();        int ttl = wl * l.size();        //find        for (int start = 0; start < wl; start++)        {            map<string, int>temp(wordMap);            int l = start;            for (int r = start; r + wl <= s.size(); r+= wl)            {                string curStr = s.substr(r, wl);                if (temp.find(curStr) != temp.end())                {                    temp[curStr]--;                    if (temp[curStr] == 0 && r + wl - l == ttl)                    {                        res.push_back(l);                        temp[s.substr(l, wl)]++;                        l += wl;                    }                    while (temp[curStr] < 0)                    {                        temp[s.substr(l, wl)]++;                        l += wl;                    }                }                else                {                    l = r + wl;                    temp = wordMap;                    if (l + ttl > s.size()) break;                }            }        }        return res;    }