Question

所以我正在完成一项任务，而且我很困难。这个问题是http://www.cs.princeton.edu/courses/archive/fall05/cos226/assignments/lines.html 基本上它涉及写3个班级。第一个，point.java用于保存关于XY平面中的点的信息。然后分配要求您在同一条线上找到4个点的每个组合（具有相同的斜率）首先使用Brute.java进行比较，然后比较每个可能的组合，然后使用Fast.jave进行比较，在那里选择起点并从那里进行比较，然后再不使用该点。这样你就可以避免检查相同的点。

我坚持的部分是最后一部分，Fast.java。我得到几乎正确的输出，除了最后两行被交换的事实。鉴于输入

15
10000 0
8000 2000
2000 8000
0  10000

20000 0
18000 2000
2000 18000

10000 20000
30000 0
0 30000
20000 10000

13000 0
11000 3000
5000 12000
9000 6000

我在找到斜坡上的点之前对数组进行排序。在调用getLines之前，已排序的数组如下所示：

(10000, 0)
(8000, 2000)
(2000, 8000)
(0, 10000)
(20000, 0)
(18000, 2000)
(2000, 18000)
(10000, 20000)
(30000, 0)
(0, 30000)
(20000, 10000)
(13000, 0)
(11000, 3000)
(5000, 12000)
(9000, 6000)

这是我的Point类：

  public class Point implements Comparable<Point> {

public final int x, y;

// compare points by slope
public final Comparator<Point> SLOPE_ORDER = new Comparator<Point>()
        {
            @Override
            public int compare(Point arg0, Point arg1) {
                // TODO Auto-generated method stub
                if(slopeTo(arg0) < slopeTo(arg1)){
                    return -1;
                }
                if(slopeTo(arg0) > slopeTo(arg1)){
                    return 1;
                }
                else{
                    return 0;
                }
            }
        };

// create the point (x, y)
public Point(int x, int y) {
    this.x = x;
    this.y = y;
}

// plot this point to standard drawing
public void draw() {
    /* DO NOT MODIFY */
    StdDraw.point(x, y);
}

// draw line between this point and that point to standard drawing
public void drawTo(Point that) {
    /* DO NOT MODIFY */
    StdDraw.line(this.x, this.y, that.x, that.y);
}

// slope between this point and that point
public double slopeTo(Point that) {
    // TODO: Implement this

    if(this.x == that.x){
        return Double.NEGATIVE_INFINITY;
    }
    else if(that.y - this.y == 0){
        return 0;
    }
    else if(this.x == that.x){
        return Double.POSITIVE_INFINITY;
    }
    else{
        return (((double)that.y - (double)this.y)/((double)that.x - (double)this.x));
    }
}

/**
 * Is this point lexicographically smaller than that one? comparing
 * y-coordinates and breaking ties by x-coordinates
 */
public int compareTo(Point that) {
    // TODO: Implement this
    if(this.y < that.y){
        return -1;
    }
    else if(this.y > that.y){
        return 1;
    }
    else{
        if(this.x < that.x){
            return -1;
        }
        else if(this.x > that.x){
            return 1;
        }
        else{
            return 0;
        }
    }
}

这是我的fast.java类：

private static void getLines(Point[] points) {
   Point p = points[0];
   Point[] lines = new Point[points.length];
   lines[0] = p;
   int pointsOnLine = 0;
   double lastSlope = p.slopeTo(points[1]);
   for(int i = 1; i < points.length; i++) {
       Point newPoint = points[i];
       double slope = p.slopeTo(newPoint);
       if(slope == lastSlope) {
           pointsOnLine++;
           lines[pointsOnLine] = newPoint;
       }
       else {
           if(pointsOnLine >= 3) {
               printLine(lines, pointsOnLine + 1);
           }
           pointsOnLine = 1;
           lines[1] = newPoint;
       }
       lastSlope = slope;
   }

   if(pointsOnLine >= 3) {
       printLine(lines, pointsOnLine + 1);
   }
}

private static void printLine(Point[] lines, int size) {
   Arrays.sort(lines, 1, size);
   if(lines[0].compareTo(lines[1]) < 0) {
       StdOut.print(lines[0]);
       for(int i = 1; i < size; i++) {
           StdOut.print(" -> " + lines[i]);
       }
       StdOut.println();
   }
}

我的主要功能

public static void main(String[] args) {
   In in = new In();
   int N = in.readInt();
   Point[] points = new Point[N];
   for(int i=0; i< N; i++) {
       int x = in.readInt();
       int y = in.readInt();
       points[i] = new Point(x,y);
   }

   in.close();
   Point [] pointsCopy = Arrays.copyOf(points, points.length);
   for (Point originPoint : points) {
       Arrays.sort(pointsCopy, originPoint.SLOPE_ORDER);
       getLines(pointsCopy);
   }
}

我的节目输出

(10000, 0) -> (8000, 2000) -> (2000, 8000) -> (0, 10000)
(10000, 0) -> (13000, 0) -> (20000, 0) -> (30000, 0)
(30000, 0) -> (20000, 10000) -> (10000, 20000) -> (0, 30000)
(13000, 0) -> (11000, 3000) -> (9000, 6000) -> (5000, 12000)

正确的输出是

(10000, 0) -> (8000, 2000) -> (2000, 8000) -> (0, 10000)
(10000, 0) -> (13000, 0) -> (20000, 0) -> (30000, 0)
(13000, 0) -> (11000, 3000) -> (9000, 6000) -> (5000, 12000)
(30000, 0) -> (20000, 10000) -> (10000, 20000) -> (0, 30000)

任何可以看到我出错的地方的人？感谢。

Answer 1

输出将遵循pointsCopy数组中的顺序。 Arrays.sort正在以导致输出的方式排序。您可以重新排列pointsCopy数组中的元素或更改数组排序逻辑以获得所需的输出。

通过排序Java

1 个答案: