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.)

class Solution {
    // Approach 1
    public int[][] kClosest2(int[][] points, int K) {
        Queue<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) {
            K--;
            res[K] = pq.poll();
        }
        return res;
    }

    // Approach 2
    public int[][] kClosest(int[][] points, int K) {
        int len = points.length, l = 0, r = len - 1;
        while (l <= r) {
            int mid = partition(points, l, r);
            if (mid == K) {
                break;
            } else if (mid < K) {
                l = mid + 1;
            } else {
                r = mid - 1;
            }
        }
        return Arrays.copyOfRange(points, 0, K);
    }

    private int compare(int[] p1, int[] p2) {
        return p1[0] * p1[0] + p1[1] * p1[1] - p2[0] * p2[0] - p2[1] * p2[1];
    }

    private int partition(int[][] A, int start, int end) {
        int p = start;
        for (int i = start; i <= end - 1; i++) {
            if (compare(A[i], A[end]) < 0) {
                swap(A, p, i);
                p++;
            }
        }
        swap(A, p, end);
        return p;
    }

    private void swap(int[][] nums, int i, int j) {
        int[] temp = nums[i];
        nums[i] = nums[j];
        nums[j] = temp;
    }
}