OA原题描述参考：

注意：具体抽到的题目可能会有细微变化。

亚麻建了个新warehouse，从多个地点运东西过去，选择最近的k个点。就是k closest points，用priorityqueue。

这题类似如下题目。

Given somepointsand a pointoriginin two dimensional space, findkpoints out of the some points which are nearest toorigin.
Return these points sorted by distance, if they are same with distance, sorted by x-axis, otherwise sorted by y-axis.

Example

Given points =[[4,6],[4,7],[4,4],[2,5],[1,1]], origin =[0, 0], k =3
return[[1,1],[2,5],[4,4]]

解题思路参考：

解决办法就是heap（priorityqueue），坑：要自己写comparator，注意比较距离的公式是x*x+y*y.

参考答案：

/**
 * Definition for a point.
 * class Point {
 *     int x;
 *     int y;
 *     Point() { x = 0; y = 0; }
 *     Point(int a, int b) { x = a; y = b; }
 * }
 */
public class Solution {
    /**
     * @param points a list of points
     * @param origin a point
     * @param k an integer
     * @return the k closest points
     */
    private Point global_origin = null;
    public Point[] kClosest(Point[] points, Point origin, int k) {
        // Write your code here
        global_origin = origin;
        PriorityQueue<Point> pq = new PriorityQueue<Point> (k, new Comparator<Point> () {
            @Override
            public int compare(Point a, Point b) {
                int diff = getDistance(b, global_origin) - getDistance(a, global_origin);
                if (diff == 0)
                    diff = b.x - a.x;
                if (diff == 0)
                    diff = b.y - a.y;
                return diff;
            }
        });

        for (int i = 0; i < points.length; i++) {
            pq.offer(points[i]);
            if (pq.size() > k)
                pq.poll();
        }

        k = pq.size();
        Point[] ret = new Point[k];  
        while (!pq.isEmpty())
            ret[--k] = pq.poll();
        return ret;
    }

    private int getDistance(Point a, Point b) {
        return (a.x - b.x) * (a.x - b.x) + (a.y - b.y) * (a.y - b.y);
    }
}