Return 如果点位于矩形边界内则为真
Return true if the point lies within the boundaries of the rectangle
我需要找到位于矩形边界或其边界内的点。到目前为止，我的代码无法确定该点是否位于矩形的边界或其边界内。请帮忙。
这是我目前所掌握的，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;
}
这是我的其余代码：
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.
}
}
下面是执行 Contains 方法的代码。如果找到该点，应该给我 15 分，但现在它没有给我任何分数。
回答正确会有加分哦！提前致谢。
@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)));
}
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;
}
或者，如评论中所述，您可以使用 IsBetween
public static boolean contains(Rectangle r, Point p)
{
return Assignment6.isBetween(p.x,r.x,r.x+r.width) && Assignment6.isBetween(p.y,r.y,r.y+r.height);
}
我需要找到位于矩形边界或其边界内的点。到目前为止，我的代码无法确定该点是否位于矩形的边界或其边界内。请帮忙。
这是我目前所掌握的，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;
}
这是我的其余代码：
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.
}
}
下面是执行 Contains 方法的代码。如果找到该点，应该给我 15 分，但现在它没有给我任何分数。 回答正确会有加分哦！提前致谢。
@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)));
}
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;
}
或者，如评论中所述，您可以使用 IsBetween
public static boolean contains(Rectangle r, Point p)
{
return Assignment6.isBetween(p.x,r.x,r.x+r.width) && Assignment6.isBetween(p.y,r.y,r.y+r.height);
}