We have a list of points
on the plane. Find the K
closest points to the origin (0, 0)
.
(Here, the distance between two points on a plane is the Euclidean distance.)
You may return the answer in any order. The answer is guaranteed to be unique (except for the order that it is in.)
Example 1:
Input: points = [[1,3],[-2,2]], K = 1
Output: [[-2,2]]
Explanation:
The distance between (1, 3) and the origin is sqrt(10).
The distance between (-2, 2) and the origin is sqrt(8).
Since sqrt(8) < sqrt(10), (-2, 2) is closer to the origin.
We only want the closest K = 1 points from the origin, so the answer is just [[-2,2]].
Example 2:
Input: points = [[3,3],[5,-1],[-2,4]], K = 2
Output: [[3,3],[-2,4]]
(The answer [[-2,4],[3,3]] would also be accepted.)
class Solution { public int[][] kClosest(int[][] points, int K) { // keep max heap PriorityQueue<Cell> pq = new PriorityQueue<>((a, b) -> (b.x)*(b.x)+ (b.y)*(b.y) - (a.x)*(a.x) - (a.y)*(a.y)); for (int[] point: points) { pq.offer(new Cell(point[0], point[1])); if (pq.size() > K) { pq.poll(); } } int[][] res = new int[K][2]; for (int i = K - 1; i >= 0; i--) { Cell cur = pq.poll(); res[i][0] = cur.x; res[i][1] = cur.y; } return res; } } class Cell { int x; int y; public Cell(int x, int y) { this.x = x; this.y = y; } }