【ATT】Substring with Concatenation of All Words

 

 

和minimum window string的原理一样。维护一个window

 

    vector<int> findSubstring(string S, vector<string> &L) {
        // Start typing your C/C++ solution below
        // DO NOT write int main() function
        vector<int> res;
        if(S.empty()||L.empty())
            return res;
        unordered_map<string,int> dict;
        vector<int> needed(L.size(),0);
        int i,j;
        //precess
        for(i=0;i<L.size();i++)
        {
            if(dict.find(L[i])==dict.end())
            {
                dict[L[i]] = i;   //先对L中每个字符串编号。
                needed[i] = 1;    //统计每个字符串出现的次数。
            }else
                needed[dict[L[i]]]++;
        }
        
        int sLen = S.size();
        int LLen = L.size();
        int len = L[0].size();
        
        vector<int> pos(sLen,-1);  //pos[i]==-1,表示[i,i+len-1]组成的字符串不在L中，否则，存储其编号
        
        for(i=0;i<=sLen-len;i++)
        {
            string str = S.substr(i,len);
            if(dict.find(str)!=dict.end())
                pos[i] = dict[str];
        }
        
        for(int offset = 0;offset<len;offset++)
        {
            int count = 0;
            vector<int> found(LLen,0);  //统计当前window发现的L中每个字符串个数
            int begin = offset,end = offset;
            while(end<=sLen-len)
            {
                if(pos[end]==-1)  //cur str not in L
                {
                    end += len;
                    begin = end;
                    found.clear(); //清空
                    found.resize(LLen,0);
                    count = 0;
                }else
                {
                    found[pos[end]]++;
                    if(found[pos[end]]<=needed[pos[end]])
                        count++;   //统计cnt
                    
                    if((end-begin)/len+1==LLen)
                    {
                        if(count==LLen)   //valid window
                            res.push_back(begin);
                        
                        if(found[pos[begin]]<=needed[pos[begin]])
                            count--;
                        found[pos[begin]]--;
                        begin+=len;
                    }
                    end+=len;
                }
            }
        }
        
        return res;
        
        
    }