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.)
题意求离原点距离最近的K个点
思路
维护一个minHeap
代码
1 class Solution { 2 public int[][] kClosest(int[][] points, int K) { 3 int[][] res = new int[K][2]; 4 5 PriorityQueue<int[]> minHeap = new PriorityQueue<>((a, b) -> (a[0] * a[0] + a[1] * a[1]) - (b[0] * b[0] + b[1] * b[1])); 6 for (int[] point : points) { 7 minHeap.add(point); 8 } 9 10 for (int i = 0; i < K; i++) { 11 res[i] = minHeap.poll(); 12 } 13 return res; 14 15 } 16 }
转载于:https://www.cnblogs.com/liuliu5151/p/11124118.html
