LeetCode--No.973--K Closest Points to Origin

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

知道PriorityQueue的存在，但是被我給忘了。抄的答案代碼。
今天要再看一下priorityqueue的comparator定義等等

class Solution {
    public int[][] kClosest(int[][] points, int K) {
        PriorityQueue<int[]> pq = new PriorityQueue<int[]>((p1, p2) -> p2[0]*p2[0] + p2[1] * p2[1] - p1[0] * p1[0] - p1[1] * p1[1]);
        for(int[] p : points){
            pq.offer(p);
            if (pq.size() > K){
                pq.poll();
            }
        }
        int[][] res = new int[K][2];
        while( K > 0){
            res[--K] = pq.poll();
        }
        return res;
    }
}