我需要找到点位于矩形或其边界的边界内。到目前为止我的代码并没有找到该点是否位于矩形或其边界的边界内。请帮忙。
这是我到目前为止所做的，这是唯一没有正确执行的部分。
public static boolean contains(Rectangle r, Point p)
{
// TODO - you need to implement this. May want to use isInBetween
return p.x >= r.x && p.y >= r.y && p.x < r.x + r.width && p.y < r.y + r.height;
}
以下是我的其余代码：
package assignment;
import java.awt.Point;
import java.awt.Rectangle;
public class Assignment6 {
//this is just sample code demonstrating aliasing
public static void aliasing1()
{
Point pt1=new Point(10,10);
Point pt2=new Point(5,5);
Point alias=pt1;
alias.x+=pt2.x;
alias=pt2;
alias.y+=5;
int sumX=pt1.x+pt2.x;
int sumY=pt1.y+pt2.y;
System.out.println("sumX= "+sumX+ " -sumY= "+sumY);
}
// more sample code
public static void playWithRectangles(){
Rectangle r=new Rectangle(10, 10, 5, 7);
System.out.println(r.x);
}
/* Examples */
/*
* A point is to the left of another if its x coordinate is less than the other
*/
public static boolean isToTheLeft(Point p1, Point p2)
{
return p1.x < p2.x;
}
/*
* A point is to the right of another if its x coordinate is bigger than the other
*/
public static boolean isToTheRight(Point p1, Point p2)
{
// TODO - you need to implement this
return p1.x > p2.x;
}
/*
* We use 'screen coordinates', with y=0 being at the top of the screen and growing downwards
* so a point is above another if its y coordinate is less
*/
public static boolean isAbove(Point p1, Point p2)
{
return p1.y<p2.y;
}
/* isBelow returns true if the first point is below the second one (a point is below another if its y coordinate is bigger)
*/
public static boolean isBelow(Point p1, Point p2)
{
// TODO - you need to implement this
return p1.y>p2.y;
}
// Example - calculate area of a rectangle
public static int getArea(Rectangle r)
{
return r.width*r.height;
}
// Example - grow a rectangle's width and height by a factor. Notice we're modifying the rectangle
public static void growRectangle(Rectangle r, int factor)
{
r.width *= factor;
r.height *= factor;
}
/* The rectangle has the top-left coordinates (in screen coordinates, y=0 is top, y grows down)
* So the bottom-right coordinate can be calculated by adding the width and height to x and y, respectively
*/
public Point getBottomRightCorner(Rectangle r)
{
return new Point(r.x+r.width, r.y+r.height);
}
/*
* Returns the center of the rectangle. The center is obtained by adding half the width and half the height to the x and y coordinates respectively
* Round DOWN (if needed) when calculating the center.
*/
public static Point getCenter(Rectangle r)
{
return new Point(r.x+r.width/2, r.y+r.height/2);
}
/**
* Example Returns true if x is between low and high (inclusive)
*/
public static boolean isInBetween(int x, int low, int high)
{
return x>=low && x<=high;
}
/**
* Returns true if the rectangle contains the point; that is, if the point lies within the boundaries of the rectangle
* If the point is exactly in the border we still return true.
* @param r - the rectangle
* @param p - the point
* @return true if the point lies within the rectangle or on its border
*/
public static boolean contains(Rectangle r, Point p)
{
// TODO - you need to implement this. May want to use isInBetween
return p.x >= r.x && p.y >= r.y && p.x < r.x + r.width && p.y < r.y + r.height;
}
/* Person example */
public static void usingPerson() {
Person p1=new Person("Orlando","Karam",40); // we construct with first, last names, age
System.out.println(p1.getFirstName()); // would print Orlando
System.out.println(p1.getLastName()); // would print Karam
System.out.println(p1.getAge()); // would print 40
}
/* Example - two people are relatives if they have the same last name :) */
public boolean areRelatives(Person p1, Person p2)
{
return p1.getLastName().equals(p2.getLastName());
}
/* Example creating a description for a person, with name and age*/
public String getDescription(Person p)
{
return p.getFirstName()+" "+p.getLastName()+" is "+p.getAge()+" years old";
}
/* return true if the persons' first and last name and age are equal */
public static boolean personEquals(Person p1, Person p2)
{
// TODO - you need to implement this
return p1.getFirstName().equals(p2.getFirstName()) && p1.getLastName().equals(p2.getLastName());
}
/* returns a person's full name, composed of the first name, a space and the last name, as in
* Orlando Karam
*/
public static String getFullName(Person p)
{
// TODO - you need to implement this
return p.getFirstName()+" "+p.getLastName();
}
/* returns a 'formal' full name, composed of the last name, a comma, a space and the first name, as in
* Karam, Orlando
*/
public static String getFormalFullName(Person p)
{
// TODO - you need to implement this
return p.getLastName()+","+" "+p.getFirstName();
}
public static void main(String[] args) {
// you can use this as you wish to test or exercise your function. Not graded.
}
}
以下是执行Contains方法的代码。如果发现这一点，假设给我15分，但现在它没有给我任何积分。 积分将被授予正确的答案！提前谢谢。
@Grade(points = 15)
@Test
public void testContains() {
Assert.assertTrue(Assignment6.contains(new Rectangle(0, 0, 10, 10), new Point(5, 5)));
Assert.assertTrue(Assignment6.contains(new Rectangle(0, 0, 10, 10), new Point(0, 0)));
Assert.assertTrue(Assignment6.contains(new Rectangle(0, 0, 10, 10), new Point(0, 10)));
Assert.assertFalse(Assignment6.contains(new Rectangle(0, 0, 10, 10), new Point(11, 5)));
Assert.assertFalse(Assignment6.contains(new Rectangle(0, 0, 10, 10), new Point(5, 15)));
}
答案 0 :(得分：0)
def inorder_keys(self):
if self.head.left is not None:
yield from self.head.left.inorder_keys()
yield self.head_key
if self.head.right is not None:
yield from self.head.right.inorder_keys()
或者，如评论中所述，您可以使用IsBetween
public static boolean contains(Rectangle r, Point p)
{
// TODO - you need to implement this. May want to use isInBetween
return p.x >= r.x && p.y >= r.y && p.x <= r.x + r.width && p.y < =r.y + r.height;
}