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

 My answer: use max heap to keep track of smallest K elements, poll larger element every time when size is over K. O(N logk)

Other solutions can be quickselect,

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class Solution { class Point { public int i; public int j; public double dist; public Point (int i, int j) { this.i = i; this.j = j; this.dist = Math.sqrt(i*i + j*j); } } public int[][] kClosest(int[][] points, int K) { if (points.length <= K) { return points; } // max heap PriorityQueue<Point> heap = new PriorityQueue<Point>((b, a) -> {return Double.compare(a.dist, b.dist);}); for (int[] rawPoint : points) { Point point = new Point(rawPoint[0], rawPoint[1]); heap.add(point); if (heap.size() > K) { heap.poll(); } } int[][] results = new int[K][2]; for (int i = 0; i < K; i++) { Point point = heap.poll(); results[i] = new int[]{point.i, point.j}; } return results; } }