K Closest Points

寻找最近的K个点

最新推荐文章于 2020-11-09 09:40:37 发布

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本文介绍了一种算法，用于从二维空间中找到距离给定点原点最近的K个点，并按照距离、X轴和Y轴顺序进行排序。示例中使用了Java实现，通过优先队列和自定义比较器来高效解决问题。

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.

Have you met this question in a real interview?

Yes

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]]

java

/**
 * 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
     */
    public Point[] kClosest(Point[] points, Point origin, int k) {
        // write your code here
        Comparator<Point> cmp = new Comparator<Point>() {
            public int compare(Point A, Point B) {
                int ad = getDistance(A.x, origin.x) + getDistance(A.y, origin.y);
                int bd = getDistance(B.x, origin.x) + getDistance(B.y, origin.y);
                if (ad == bd) {
                    int axd = A.x - origin.x;
                    int bxd = B.x - origin.x;
                    if (axd == bxd) {
                        int ayd = A.y - origin.y;
                        int byd = B.y - origin.y;
                        return -(ayd - byd);
                    } else {
                        return -(axd - bxd);
                    }
                } else {
                    return -(ad - bd);    
                }
            }  
        };
        Queue<Point> queue = new PriorityQueue<>(k + 1, cmp);
        for (int i = 0; i < points.length; i++) {
            queue.offer(points[i]);
            if (queue.size() > k) {
                queue.poll();
            }
        }
        Point[] arr = new Point[k];
        for (int i = k -1; i >= 0; i--) {
            arr[i] = queue.poll();
        }
        return arr;
    }
    private int getDistance(int A, int B) {
        int val = (A - B) * (A - B);
        return val;
    }
}