[LeetCode] 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.

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.

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.

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

统计一个圆中点的数目。

给你一个数组 points ，其中 points[i] = [xi, yi] ，表示第 i 个点在二维平面上的坐标。多个点可能会有 相同 的坐标。

同时给你一个数组 queries ，其中 queries[j] = [xj, yj, rj] ，表示一个圆心在 (xj, yj) 且半径为 rj 的圆。

对于每一个查询 queries[j] ，计算在第 j 个圆 内 点的数目。如果一个点在圆的 边界上 ，我们同样认为它在圆 内 。

请你返回一个数组 answer ，其中 answer[j]是第 j 个查询的答案。

来源：力扣（LeetCode）
链接：https://leetcode.cn/problems/queries-on-number-of-points-inside-a-circle
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这是一道数学题，这里我提供一个暴力的思路。

题目给了每个圆心的坐标和半径，那么判断一个坐标是否在某个圆里面，我们就判断这个坐标到圆心的距离是否 <= 这个圆的半径即可。

时间O(mn)

空间O(n) - output array

Java实现

class Solution {
    public int[] countPoints(int[][] points, int[][] queries) {
        int n = queries.length;
        int[] res = new int[n];
        for (int i = 0; i < n; i++) {
            int x = queries[i][0];
            int y = queries[i][1];
            int r = queries[i][2] * queries[i][2];
            for (int[] p : points) {
                res[i] += (p[0] - x) * (p[0] - x) + (p[1] - y) * (p[1] - y) <= r ? 1 : 0;
            }
        }
        return res;
    }
}

 

LeetCode 题目总结