0
我有一個包含一堆代碼的細分類。我堅持使用兩種方法，isParallel和縮短方法。這是到目前爲止我的代碼（我有一個點類，以及其鏈接到這個類）：如何在段類中使用isParallel並縮短方法？
public class Segment {
\t //two Points that hold endpoints of the segment
\t private Point p1, p2;
\t
\t
\t //Default constructor that will set the endpoints to new
\t //Points with values (0,0) and (4,4)
\t public Segment(){
\t \t this(0, 0, 4, 4);
\t }
\t
\t //Parameterized constructor that accepts (int x1, int y1, int x2, int y2)
\t //and creates and sets the endpoints
\t public Segment(int x1, int x2, int y1, int y2){
\t \t this.p1 = new Point(x1, y1);
\t \t this.p2 = new Point(x2, y2);
\t }
\t
\t //Parameterized constructor that accepts (Point p1, Point p2) and sets both
\t //the endpoints to a deep copy of the Points that are passed in.
\t public Segment(Point p1, Point p2){
\t \t this.p1 = new Point(p1.getX(), p1.getY());
\t \t this.p2 = new Point(p2.getX(), p2.getY());
\t }
\t
\t //Copy constructor that accepts a Segment and initializes the data (of the
\t //new Segment being created) to be the same as the Segment that was received.
\t public Segment(Segment other){
\t \t p1 = other.getP1();
\t \t p2 = other.getP2();
\t }
\t public Point getP1(){
\t \t return p1;
\t }
\t
\t public Point getP2(){
\t \t return p2;
\t }
\t
\t //The length method returns the length of the Segment.In fact, this method is same as distanceTo method
\t //So we can use distanceTo method which is already defined in Point class
\t public double length(){
\t \t return (p1.distanceTo(p2));
\t }
\t
\t //The translate method returns nothing and should translate, or shift, \t
\t //itself (the Segment) by the distances passed in
\t public void translate(int xmove, int ymove) {
\t p1.translate(xmove,ymove);
\t p2.translate(xmove,ymove);
\t }
\t
\t //The midpoint method calculates and returns the midpoint of the Segment as a new Point
\t public Point midpoint(){
\t \t return (p1.halfWayTo(p2)); \t
\t }
\t
\t //The slope method returns the slope of the Segment as a double.
\t public double slope(){
\t \t \t return (double)(p2.getY() - p1.getY())/(p1.getX() - p2.getX());
\t }
\t /**
\t * The isParallel method returns true/false depending on whether the current Segment
\t * is parallel to the Segment received. Think about how you can tell if two segments
\t * are parallel. Note: Two overlapping segments ARE parallel.
\t */
\t public boolean isParallel(Segment s1){
{
/**
\t * The shorten method changes its (the Segment's) endpoints so that they are both halfway
\t * to the midpoint. Example: The segment (0,0)---(12,16) has midpoint (6,8). After
\t * calling the shorten method, the segment should be (3,4)---(9,12). Each endpoint
\t * has moved in toward the midpoint (which stayed the same). So (3,4) is halfway between
\t * (0,0) and (6,8) and (9,12) is halfway between (12,16) and (6,8).
\t */
public void shorten();有人可以給我如何在我的代碼中使用這兩種方法的想法。我非常感謝你的幫助。
謝謝！
我的直覺是你不問你想問什麼。你想實現'isParallel'和'shorten'嗎？ – 2014-09-26 23:07:56
我只需要知道如何編寫這些方法！或者，也許是！我想知道如何實施它們！ – Hr0419 2014-09-27 02:13:45
事實上，你的問題的重寫是由於！至於手頭的事情，你所需要的只是對算法的一些數學支持。 – 2014-09-27 06:48:06