K Closest Points

Given (0,0) as the centre and an array containing pair of coordinate such that 

(1,1)

(-2,4)

(3,4)

(-11,5)

...

(x,y).

If ask you return k closest points, how would you do it?

import java.util.Comparator;import java.util.PriorityQueue;public class kclosestPoints {public static void main(String[] args){kclosestPoints kp = new kclosestPoints();CPoint c1 = new CPoint(1,1);CPoint c2 = new CPoint(-7,10);CPoint c3 = new CPoint(5,3);CPoint c4 = new CPoint(4,3);CPoint c5 = new CPoint(11,2);CPoint c6 = new CPoint(-14,-20);CPoint[] input = {c1,c2,c3,c4,c5,c6};for(CPoint k : kp.findClose(input, 3)){System.out.println(k.x+", "+k.y);}}public CPoint[] findClose(CPoint[] arr, int k){if(k==0)return new CPoint[0];Comparator<CPoint> cmp = new Comparator<CPoint>(){@Overridepublic int compare(CPoint c1, CPoint c2) {// TODO Auto-generated method stubdouble d1 = c1.x * c1.x + c1.y*c1.y;double d2 = c2.x * c2.x + c2.y*c2.y;return  (int) (d1-d2);}};//k is capacity, cmp is an obj implementing Comparator, in which define the compare rulesPriorityQueue<CPoint> heap = new PriorityQueue<CPoint>(k,  cmp);for(CPoint p : arr){heap.offer(p);  //offer is to input }CPoint[] res = new CPoint[k];for(int i=0;i<k;i++){res[i] = heap.poll(); //poll first k prior elements}return res;}static class CPoint{double x;double y;public CPoint(double a, double b){this.x = a;this.y = b;}}}

这道题目是非常好的联系priorityqueue的机会，一般来说编程的顺序就是

先new一个cmp对象from Comparator接口，匿名类直接overwrite 里面的compare方法，里面复写自己compare方法。

然后new priorityqueue同时后面直接根据constructor new这个heap，两个参数，k代表capacity，cmp就是刚才new的comparator。