程序媛詹妮弗
终身学习

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 }

 

posted on 2019-07-03 02:34  程序媛詹妮弗  阅读(257)  评论(0编辑  收藏  举报