[Google] LeetCode 2013 Detect Squares

You are given a stream of points on the X-Y plane. Design an algorithm that:

• Adds new points from the stream into a data structure. Duplicate points are allowed and should be treated as different points.
• Given a query point, counts the number of ways to choose three points from the data structure such that the three points and the query point form an axis-aligned square with positive area.
• An axis-aligned square is a square whose edges are all the same length and are either parallel or perpendicular to the x-axis and y-axis.

Implement the DetectSquares class:

• DetectSquares() Initializes the object with an empty data structure.
• void add(int[] point) Adds a new point point = [x, y] to the data structure.
• int count(int[] point) Counts the number of ways to form axis-aligned squares with point point = [x, y] as described above.

Solution

每次查询时，给定一个点，我们要找出能构成正方形的数量。注意到，对于该点，那么另一个端点应该有相同的 $x$ 或者有相同的 $y$. 不妨固定 $x$，找出 $x$ 相同的所有 $y$， 做差便得到可能的边长，然后检查另外两个点是否存在。

这里使用 $map$ 套 $map$ 来表示一个点被记录了多少次

点击查看代码

class DetectSquares {
private:
    unordered_map<int, unordered_map<int,int>> dots;
public:
    DetectSquares() {
        
    }
    
    void add(vector<int> point) {
        int x = point[0], y = point[1];
        dots[x][y]++;
    }
    
    int count(vector<int> point) {
        int ans = 0;
        int x1 = point[0], y1 = point[1];
        if(dots.find(x1)==dots.end())return 0;
        for(auto ele: dots[x1]){
            int y2 = ele.first;
            if(y2==y1)continue;
            int dis = abs(y2-y1);
            
            // case 1
            int x2 = x1+dis;
            if(dots.find(x2)!=dots.end() && dots[x2].find(y1)!=dots[x2].end() && dots[x2].find(y2)!=dots[x2].end()){
                ans+= dots[x2][y1]*dots[x2][y2]*ele.second;
            }
            // case 2
            x2 = x1-dis;
            if(dots.find(x2)!=dots.end() && dots[x2].find(y1)!=dots[x2].end() && dots[x2].find(y2)!=dots[x2].end()){
                ans+= dots[x2][y1]*dots[x2][y2]*ele.second;
            }
        }
        return ans;
    }
};
 
/**
 * Your DetectSquares object will be instantiated and called as such:
 * DetectSquares* obj = new DetectSquares();
 * obj->add(point);
 * int param_2 = obj->count(point);
 */