1828. Queries on Number of Points Inside a Circle

You are given an array points where points[i] = [xi, yi] is the coordinates of the ith point on a 2D plane. Multiple points can have the same coordinates.

You are also given an array queries where queries[j] = [xj, yj, rj] describes a circle centered at (xj, yj) with a radius of rj.

For each query queries[j], compute the number of points inside the jth circle. Points on the border of the circle are considered inside.

Return an array answer, where answer[j] is the answer to the jth query.

 

 

Constraints:

• 1 <= points.length <= 500
• points[i].length == 2
• 0 <= x​​​​​​i, y​​​​​​i <= 500
• 1 <= queries.length <= 500
• queries[j].length == 3
• 0 <= xj, yj <= 500
• 1 <= rj <= 500
• All coordinates are integers.

 

Example 1:

Input: points = [[1,3],[3,3],[5,3],[2,2]], queries = [[2,3,1],[4,3,1],[1,1,2]]
Output: [3,2,2]
Explanation: The points and circles are shown above.
queries[0] is the green circle, queries[1] is the red circle, and queries[2] is the blue circle.

Example 2:

Input: points = [[1,1],[2,2],[3,3],[4,4],[5,5]], queries = [[1,2,2],[2,2,2],[4,3,2],[4,3,3]]
Output: [2,3,2,4]
Explanation: The points and circles are shown above.
queries[0] is green, queries[1] is red, queries[2] is blue, and queries[3] is purple.

Follow up: Could you find the answer for each query in better complexity than O(n)?

判断一个点是否在圆上怎么判断呢？？？

 

点到圆心距离的平方<=半径的平方

 C++

class Solution {
public:
    vector<int> countPoints(vector<vector<int>>& points, vector<vector<int>>& queries) {
        vector<int> res(queries.size());
        for(int i=0;i<queries.size();i++){
            int x=queries[i][0],y=queries[i][1],r=queries[i][2];
            for(int j=0;j<points.size();j++){
                if((x-points[j][0])*(x-points[j][0])+(y-points[j][1])*(y-points[j][1])<=r*r)
                    res[i]++;
                    
            }
           
        }
        return res;
    }
};