标签：answer col tee for input exce style The ble
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：
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 }
LeetCode 973. K Closest Points to Origin
标签：answer col tee for input exce style The ble
原文地址：https://www.cnblogs.com/Dylan-Java-NYC/p/12216495.html
