leetcode 1828. Queries on Number of Points Inside a Circle（python）

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描述

You are given an array points where points[i] = [x_i, y_i] is the coordinates of the i^th point on a 2D plane. Multiple points can have the same coordinates.

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

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

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

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.

Note:

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

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

解析

根据题意，就是给出一个点列表 points ，然后又给出一组圆列表，每个圆元素有三个值，前两个是表示中心点为位置，第三个表示的是圆半径，计算每个每个圆可以囊括 points 中的点的个数，并且按照圆出现的顺序，在列表中显示每个圆能囊括的点的个数，其中点如果刚好在边上也算有效点。思路比较简单：

• 初始化一个结果列表 result ，长度为圆的个数
• 两层循环，一层循环点列表 points ，第二层循环圆列表 queries ，然后判断点到第 i 个圆的圆心的距离小于等于半径，即该点属于该圆的有效点，将 result[i] 加一
• 遍历结束，返回 result 即可

解答

class Solution(object):
    def countPoints(self, points, queries):
        """
        :type points: List[List[int]]
        :type queries: List[List[int]]
        :rtype: List[int]
        """
        result = [0] * len(queries)
        for point in points:
            x = point[0]
            y = point[1]
            for i,query in enumerate(queries):
                xx = query[0]
                yy = query[1]
                r = query[2]
                if abs(xx-x)**2+abs(yy-y)**2<=r**2:
                    result[i] += 1
        return result
        	      
		

运行结果

Runtime: 1160 ms, faster than 35.35% of Python online submissions for Queries on Number of Points Inside a Circle.
Memory Usage: 13.8 MB, less than 49.30% of Python online submissions for Queries on Number of Points Inside a Circle.

思考

看题目末尾的一行字：Follow up: Could you find the answer for each query in better complexity than O(n)?，这意思是要我们将时间复杂度降低到 O(n) ，这个还真有点难，相当于只能一次遍历就出结果，暂时没有思路。

原题链接：leetcode.com/problems/qu…

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