Question

我已经完成了程序的其余部分，而我只是对这些方法感到困惑。我们没有在课堂上讲这个，也没有在任何笔记上讲。我本身不想要答案，我只需要一双新鲜的眼睛。

编辑：忘记添加方法说明一个方法contains（double x，double y），如果指定点（x，y）在此矩形内，则返回true。参见下图。。一个方法contains（Rectangle 2Dangle）如果指定的矩形在此矩形内，则返回true。参见图。 .a方法重叠（矩形2D矩形），如果指定的矩形与此矩形重叠，则返回true。见图。

class Rectangle2D {
    double x;
    double y;
    double width;
    double height;


    public Boolean Contains(double x, double y) {
        return false;
    }

    public Boolean Contains(Rectangle2D R1) {
        return false;
    }

    public Boolean Overlap(Rectangle2D R1) {
        return false;
    }
}

Answer 1

一点点数学，实际上很简单。

class Rectangle2D {
    double x;
    double y;
    double width;
    double height;

    /**
     * This rectangle contains the specified point if
     *
     * The x coordinate of the point lies between x and x + width
     *
     * and
     *
     * The y coordinate of the point lies between y and y + height
     *
     * @param x - The x position of the coordinate to check
     * @param y - The y position of the coordinate to check
     * @return true if the specified coordinate lies within the rectangle.
     */
    public boolean contains(double x, double y) {
        return x >= this.x
                && y >= this.y
                && x <= this.x + this.width
                && y <= this.y + this.height;
    }

    /**
     * The rectangle contains the specified rectangle if the rectangle contains both diagonally opposite corners.
     *
     * @param r - The rectangle to check.
     * @return - true if the specified rectangle is entirely contained.
     */
    public boolean contains(Rectangle2D r) {
        return contains(r.x, r.y)
                && contains(r.x + r.width, r.y + r.height);
    }

    /**
     * The rectangle overlaps the specified rectangle if the rectangle contains any of the corners.
     *
     * @param r - The rectangle to check
     * @return - true if any corner of the rectangle is contained.
     */
    public boolean overlaps(Rectangle2D r) {
        return contains(r.x, r.y)
                || contains(r.x + r.width, r.y + r.height)
                || contains(r.x, r.y + r.height)
                || contains(r.x + r.width, r.y);
    }
}

如果一个对象重叠或位于另一个对象中，如何编写返回true的方法

1 个答案: