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.)
Note:
1 <= K <= points.length <= 10000
-10000 < points[i][0] < 10000
-10000 < points[i][1] < 10000
classSolution {publicint[][] kClosest(int[][] points,int K) {PriorityQueue<int[]> pq =newPriorityQueue<>( (a, b) -> (b[0] * b[0] + b[1] * b[1]) - (a[0] * a[0] + a[1] * a[1]));for (int[] point : points) {pq.add(point);if (pq.size() > K)pq.poll(); }int[][] ans =newint[K][];for (int i =0; i < K; i++) ans[i] =pq.poll();return ans; }}