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：

Could use maxHeap to track maximum distance, keep adding, when size > K, poll.

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

Space: O(K).

AC Java:

 1 class Solution {
 2     public int[][] kClosest(int[][] points, int K) {
 3         if(points == null || points.length < K){
 4             return points;
 5         }
 6         
 7         if(K <= 0){
 8             return new int[0][2];
 9         }
10         
11         PriorityQueue<int []> maxHeap = new PriorityQueue<>((a, b) -> b[2] - a[2]);
12         for(int [] p : points){
13             int dis = p[0] * p[0] + p[1] * p[1];
14             maxHeap.add(new int[]{p[0], p[1], dis});
15             if(maxHeap.size() > K){
16                 maxHeap.poll();
17             }
18         }
19         
20         int [][] res = new int[maxHeap.size()][2];
21         for(int i = 0; i < res.length; i++){
22             int [] cur = maxHeap.poll();
23             res[i] = new int[]{cur[0], cur[1]};
24         }
25         
26         return res;
27     }
28 }

类似Top K Frequent Elements.