标签：get put distance tps ems accept || HERE code
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 <= K <= points.length <= 10000-10000 < points[i][0] < 10000-10000 < 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 }
LeetCode 973. K Closest Points to Origin
标签：get put distance tps ems accept || HERE code
原文地址：https://www.cnblogs.com/Dylan-Java-NYC/p/10556223.html
