LeetCode 973. K Closest Points to Origin

原题链接在这里：https://leetcode.com/problems/k-closest-points-to-origin/

题目：

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. 1 <= K <= points.length <= 10000
2. -10000 < points[i][0] < 10000
3. -10000 < points[i][1] < 10000

题解：

用maxHeap来维护K shortest distence.

Time Complexity: O(nlogK). n = points.length.

Space: O(n).

AC Java:

 1 class Solution {
 2     public int[][] kClosest(int[][] points, int K) {
 3         if(points == null || points.length == 0 || K < 1){
 4             return points;
 5         }
 6
 7         PriorityQueue<int []> pq = new PriorityQueue<int []>((a, b) -> getDistanceSquare(b)-getDistanceSquare(a));
 8         for(int [] point : points){
 9             pq.add(point);
10             if(pq.size() > K){
11                 pq.poll();
12             }
13         }
14
15         int [][] res = new int[K][2];
16         for(int i = 0; i<K; i++){
17             res[i] = pq.poll();
18         }
19
20         return res;
21     }
22
23     private int getDistanceSquare(int [] point){
24         return point[0]*point[0] + point[1]*point[1];
25     }
26 }

原文地址：https://www.cnblogs.com/Dylan-Java-NYC/p/10556223.html