java编写一个测试类，要求： 创建5个点，即5个Point对象 输出X坐标最大的点的坐标。 输出点关于原点的对称点。 分别输出两点间距离最近和最远的距离。

好的，以下是我给出的答案：

import java.util.Arrays;

public class TestPoint {
    public static void main(String[] args) {
        Point[] points = new Point[5];
        points[0] = new Point(1, 2);
        points[1] = new Point(3, 4);
        points[2] = new Point(5, 6);
        points[3] = new Point(7, 8);
        points[4] = new Point(9, 10);

        // 输出X坐标最大的点的坐标
        Arrays.sort(points, (a, b) -> b.x - a.x);
        System.out.println("X坐标最大的点的坐标：" + points[0]);

        // 输出点关于原点的对称点
        for (Point point : points) {
            System.out.println(point + "关于原点的对称点：" + point.getSymmetricPoint());
        }

        // 分别输出两点间距离最近和最远的距离
        double minDistance = Double.MAX_VALUE;
        double maxDistance = Double.MIN_VALUE;
        for (int i = 0; i < points.length; i++) {
            for (int j = i + 1; j < points.length; j++) {
                double distance = points[i].getDistance(points[j]);
                if (distance < minDistance) {
                    minDistance = distance;
                }
                if (distance > maxDistance) {
                    maxDistance = distance;
                }
            }
        }
        System.out.println("两点间距离最近的距离：" + minDistance);
        System.out.println("两点间距离最远的距离：" + maxDistance);
    }
}

class Point {
    int x;
    int y;

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

    public Point getSymmetricPoint() {
        return new Point(-x, -y);
    }

    public double getDistance(Point other) {
        return Math.sqrt((x - other.x) * (x - other.x) + (y - other.y) * (y - other.y));
    }

    @Override
    public String toString() {
        return "(" + x + ", " + y + ")";
    }
}

希望我的回答能够帮到你！