Find K nearest Points

按距离的从小到大排 (2,2) (2,2) (4,4) 在PriorityQueue; 出来也是从大到小出来 因为PriorityQueue (4,4)先出来!

class Point {
    public int x;
    public int y;
    public Point(int x, int y) {
        this.x = x;
        this.y = y;
    }
   public int getX() {
      return x;
   }

  public int getY() {
      return y;
   }
}



public class Solution{
public Point[] Solution1(Point[] array, Point origin, int k){
     Point[] res = new Point[k];
     int index = 0;
     PriorityQueue<Point> pq = new PriorityQueue<Point>(k,new Comparator<Point>(){
          public int compare(Point a,Point b){
          return (int) (getDistance(b,origin)- getDistance(a,origin));
          }
     });
     for(int i =0; i<array.length;i++){
     pq.offer(array[i]);
     if(pq.size()>k){
     pq.poll();
     }
     }
     while(!pq.isEmpty()){
     res[index++] = pq.poll();
     }
   return res;
}
public double getDistance(Point a, Point b){

   return Math.sqrt((a.x-b.x)*(a.x-b.x)+(a.y-b.y)*(a.y-b.y));
}




public static void main(String args[]) {
      Solution test = new Solution();
      List<Point> list = new ArrayList<Point>();
     Point[] p = new Point[5]; 

      Point p2 = new Point(2, 2);
      Point p4 = new Point(4, 4);
      Point p3 = new Point(2, 2);
      Point p1 = new Point(6, 6);
      Point p5 = new Point(5, 5);

     p[0] = p1;
     p[1] = p2;
     p[2] = p3;
     p[3] = p4;
     p[4] = p5;
      Point origin = new Point(0, 0);

      int k = 3;
   Point[] kPoints =  test.Solution1(p,origin,3);

             for(int i = 0; i < kPoints.length; i++) { 
                    System.out.print("(" + kPoints[i].getX() + ", " + kPoints[i].getY() + ")"); 
                   if(i < kPoints.length- 1) 
              System.out.print(", "); 
}    
}}