LIKE CODING: MJ [40] Count squares

LIKE CODING: MJ [40] Count squares
Given a set of points, find all squares composed by four points of the set.

给你一些平面的点（坐标是整数），求能够成正方形的数目。
我想了想这题大概可以这么做：把每两个点之间的距离用hashmap存下来，key是
distance, values是arraylist of point pairs， 比如（p1, p2), (p3,p4)之间距离
都是6, 把他们都放到map.get(6)的list里面。这样一共要查（n,2)pairs, time也就是
(n^2). 下一步对于每一个distance, 查看对应的list size是不是大于4（至少有4个
pairs之间距离相同）。然后对于每一个list再进一步确定有没有正方形，比如
(p1,p2), (p1,p3), (p1,p4), (p2, p3), (p2, p4), (p3, p4) 都在list里，就可以确
认这四个点可以组成正方形。。。。确认的时候可以再建一个hashmap, map p1 to (p2
, p3, p4), p2 to (p3, p4), 然后找需要的pair是不是都在map里。。。就是好麻烦这
个solution, 不知有没有简单点的解法？

https://github.com/ashish1294/BugFreeCodes/blob/master/Code%20Chef/Count%20the%20squares.c

vector<vector<pair<int, int>>> findSquares(vector<pair<int, int>> &points){

    vector<vector<pair<int, int>>> ret;

    unordered_set<string> ss;//squares set

    unordered_set<string> ps;//points set

    for(auto p:points){

        string key = to_string(p.first)+" "+to_string(p.second);

        ps.insert(key);

    }

    int n = points.size();

    for(int i=0; i<n; ++i){

        int x1 = points[i].first, y1 = points[i].second;

        for(int j=i+1; j<n; ++j){

            int x2 = points[j].first, y2 = points[j].second;

            int x3 = (x1+x2+y2-y1), y3 = (y1+y2+x1-x2);

            int x4 = (x1+x2+y1-y2), y4 = (y1+y2+x2-x1);

            if(x3%2 || y3%2 || x4%2 || y4%2){continue;}

            x3/=2; y3/=2; x4/=2; y4/=2;

            string key3 = to_string(x3)+" "+to_string(y3);

            string key4 = to_string(x4)+" "+to_string(y4);

            if(ps.find(key3)!=ps.end() && ps.find(key4)!=ps.end()){

                vector<pair<int, int>> sq;

                sq.push_back(points[i]);

                sq.push_back(points[j]);

                sq.push_back(pair<int, int>(x3, y3));

                sq.push_back(pair<int, int>(x4, y4));

                sort(sq.begin(), sq.end());

                string key;

                for(auto p:sq)

                    key += to_string(p.first)+" "+to_string(p.second)+" ";

                if(ss.count(key)==0){

                    ret.push_back(sq);

                    ss.insert(key);

                }

            }

        }

    }

    return ret;

}

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