lintcode612

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]]

 

 

大根堆。O(nlogk)时间复杂度，O(k)空间复杂度。

和largest k number思路很像。只是这次要选出最近的一批，就要建大根堆。如果挤进来的东西比k多了，就踢掉一个最菜的即最远的。

 

细节：

1.实现大根堆时排序标准是距离，计算要用到Point自己本身之外的变量origin，注意因为我们写的inner class，所以要索取其他的变量必须要把它设为final。（或者你弄成全局也行，不过动不动弄全局不太好）。

2.大根堆的话就p2-p1反过来。

3.算距离的时候用了平方和。注意java里没有平方/幂次运算符，只有平方/幂次api。别傻兮兮地写x^2，^符号是异或啊亲，这是java不是matlab！用Math.pow(x, 2)，不过这个返回类型是double也要额外小心。

4.int数字做平方和小心超界限，long或double走一个。 

 

实现

/**
 * 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 globalOrigin;
    
    public Point[] kClosest(Point[] points, Point origin, int k) {
        // write your code here
        if (points == null || origin == null || k < 0 || k > points.length) {
            return new Point[0];
        }
        
        // globalOrigin = origin;
        // 确认是不是要这样？A: 是的！被内部类调用的外部变量需要设为final！或者全局。
        final Point finalOrigin = origin;
        PriorityQueue<Point> maxHeap = new PriorityQueue<Point>(new Comparator<Point>() {
            @Override
            public int compare(Point p1, Point p2) {
                long diff = distance(p2) - distance(p1);
                if (diff > 0) {
                    return 1;
                } else if (diff < 0) {
                    return -1;
                } else if (p1.x != p2.x){
                    return p2.x - p1.x;
                } else {
                    return p2.y - p1.y;
                }
            }
            
            private long distance(Point p) {
                return (long) (Math.pow(p.x - finalOrigin.x, 2) + Math.pow(p.y - finalOrigin.y, 2));
            }
        });
        
        for (Point point : points) {
            maxHeap.offer(point);
            if (maxHeap.size() > k) {
                maxHeap.poll();
            }
        }
        
        Point[] result = new Point[k];
        for (int i = result.length - 1; i >= 0; i--) {
            result[i] = maxHeap.poll();
        }
        return result;
    }
}