[LintCode] K Closest Points

Given some points and a point origin in two dimensional space, find k points out of the some points which are nearest to origin.
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]]

 

 

Solution 1. PriorityQueue with Comparable Interface 

/**
 * Definition for a point.
 * class Point {
 *     int x;
 *     int y;
 *     Point() { x = 0; y = 0; }
 *     Point(int a, int b) { x = a; y = b; }
 * }
 */
class PointWithDistance implements Comparable<PointWithDistance> {
    protected Point p;
    protected long d;
    public PointWithDistance(Point p, long d)
    {
        this.p = p;
        this.d = d;
    }
    public int compareTo(PointWithDistance b)
    {
            int ret = 0;
            if(this.d < b.d)
            {
                ret = -1;
            }
            else if(this.d > b.d)
            {
                ret =  1;
            }
            else
            {
                if(this.p.x < b.p.x)
                {
                    ret = -1;
                }
                else if(this.p.x > b.p.x)
                {
                    ret = 1;
                }
                else if(this.p.y < b.p.y)
                {
                    ret = -1;
                }
                else if(this.p.y > b.p.y)
                {
                    ret = 1;
                }
                
            }
            return ret;
    }
}
public class Solution {
    /**
     * @param points a list of points
     * @param origin a point
     * @param k an integer
     * @return the k closest points
     */
    public Point[] kClosest(Point[] points, Point origin, int k) {
        if(points == null || points.length == 0 || k <= 0)
        {
            return new Point[0];
        }
        Point[] results = new Point[k];
        PriorityQueue<PointWithDistance> pq = new PriorityQueue<PointWithDistance>(points.length);
        for(int i = 0; i < points.length; i++)
        {
            long dist = calcDist(origin, points[i]);
            pq.add(new PointWithDistance(points[i], dist));
        }
        
        for(int i = 0; i < k; i++)
        {
            results[i] = pq.poll().p;
        }
        return results;
    }
    private long calcDist(Point a, Point b)
    {
        return (a.x - b.x) * (a.x - b.x) + (a.y - b.y) * (a.y - b.y);
    }
}

 

 

Solution 2. PriorityQueue with the Comparator class

public class Solution {
    public Point[] kClosest(Point[] points, Point origin, int k) {
        if(points == null || points.length == 0 || k <= 0)
        {
            return new Point[0];
        }
        Comparator<Point> comp = new Comparator<Point>(){
            public int compare(Point p1, Point p2) {
                long dist1 = calcDist(p1, origin);
                long dist2 = calcDist(p2, origin);
                if(dist1 < dist2) {
                    return -1;
                }
                else if(dist1 > dist2) {
                    return 1;
                }
                else if(p1.x < p2.x) {
                    return -1;
                }
                else if(p1.x > p2.x) {
                    return 1;
                }
                else if(p1.y < p2.y) {
                    return -1;
                }
                else if(p1.y > p2.y) {
                    return 1;
                }
                return 0;
            }
        };
        Point[] results = new Point[k];
        PriorityQueue<Point> pq = new PriorityQueue<Point>(points.length, comp);
        for(int i = 0; i < points.length; i++)
        {
            pq.add(points[i]);
        }
        
        for(int i = 0; i < k; i++)
        {
            results[i] = pq.poll();
        }
        return results;
    }
    private long calcDist(Point a, Point b)
    {
        return (a.x - b.x) * (a.x - b.x) + (a.y - b.y) * (a.y - b.y);
    }
}

 

 

 

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