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普林斯顿大学算法Week3:CollinearPoints共线模

普林斯顿大学算法Week3:CollinearPoints共线模

作者: LittleSasuke | 来源:发表于2018-02-03 12:13 被阅读77次

    Welcome To My Blog

    总结

    (代码有详细注释)
    1. 本课讲了归并排序,作业应用是排序进行共线的模式识别,java1.8中的排序用的是tim排序,结合了归并排序与插入排序,属于稳定排序:排序之后相同元素的相对位置会不会改变
    2. Point.java中有个非常重要的方法,compareTo(),它定义:纵坐标越小则点越小,如果纵坐标相同,那么横坐标越小则点越小.(如果作业中要求横坐标也是按顺序排列,那么排序后的点集映射到二维坐标系中是非递减的折线, 这样找共线只用一层循环即可,可惜作业没加上对x的限制)
    3. 比较大小,一开始我用的是points[i-1] == point[i],尽管坐标相同但是points[i-1]不等于points[i]
      因为points[i-1]和points[i]表示引用,在堆中指向两个不同的地址,比较大小得用points[i-1].compareTo(points[i])
    4. 在FastCollinearPoints.java中,一定要注意什么时候对共线点数变量count进行判断,有两种情况,一个是,相邻元素与参考点的斜率不同;另一个是循环到最后一个元素.这两种情况在代码注释中有解释
    5. 唯一一处FAILED,扣了1分,没系统学过java,先跳过了
    Test 7: check for dependence on either compareTo() or compare()
            returning { -1, +1, 0 } instead of { negative integer,
            positive integer, zero }
      * filename = equidistant.txt
        - number of entries in student   solution: 0
        - number of entries in reference solution: 4
        - 4 missing entries in student solution, including: '(30000, 0) -> (20000, 10000) -> (10000, 20000) -> (0, 30000)'
    ==> FAILED
    
    1. week3课件中,递归调用的图示:
      这是包含两个递归调用的递归 (图示只画了一半)


      Merge.jpg

    代码

    (如需提交,请删除中文注释)

    一:Point.java

    import java.util.Comparator;
    import edu.princeton.cs.algs4.StdDraw;
    public class Point implements Comparable<Point> {
        // x-coordinate of this point
        private final int x;
        // y-coordinate of this point
        private final int y;
        // constructs the point (x, y)
        public Point(int x, int y) {
            this.x = x;
            this.y = y;
        } 
        // draws this point
        public   void draw() {
            StdDraw.point(x,y);
        }
        // draws the line segment from this point to that point
        public   void drawTo(Point that) {
            StdDraw.line(this.x, this.y, that.x, that.y);
        }
        // string representation
        public String toString() {
            return "(" + x + ", " + y + ")";
        }
        // compare two points by y-coordinates, breaking ties by x-coordinates  
        public  int compareTo(Point that) {
            if(y<that.y || (y==that.y && x<that.x)) return -1;
            else if(y==that.y && x==that.x) return 0;
            else return +1;
        }   
        // the slope between this point and that point
        public  double slopeTo(Point that) {
            if(x==that.x && y==that.y) return Double.NEGATIVE_INFINITY;
            if(x==that.x && y!=that.y) return Double.POSITIVE_INFINITY;     
            if(y==that.y) return +0.0;
            return (double)(y-that.y)/(x-that.x);
        }
        // compare two points by slopes they make with this point
        public Comparator<Point> slopeOrder(){
            return new SlopeOrder();
        }
        //nested class
        //sort rule
        private class SlopeOrder implements Comparator<Point>{
            public int compare(Point p,Point q) {
                //p点斜率大
                if(slopeTo(p)<slopeTo(q)) return -1;
                //p点斜率小
                else if(slopeTo(p)>slopeTo(q)) return 1;
                //p,q斜率相等
                else return 0;
            }
        }
        
        public static void main(String[] args) {
            Point p1 = new Point(0,0);
            Point p2 = new Point(1,1);
            Point p3 = new Point(2,2);
            Point p4 = new Point(2,1);
            Point p5 = new Point(4,1);
            System.out.println("p1.compareTo(p1) is "+p1.compareTo(p2));
            System.out.println("p2.compareTo(p1) is "+p2.compareTo(p1));
            System.out.println("p1.compareTo(p1) is "+p1.compareTo(p1)+"\n");
            
            System.out.println("p1.slopeTo(p2) is " +p1.slopeTo(p2));
            System.out.println("p1.slopeTo(p4) is "+p1.slopeTo(p4));
            System.out.println("p1.slopeTo(p1) is "+p1.slopeTo(p1));
            System.out.println("p3.slopeTo(p4) is "+p3.slopeTo(p4));
            System.out.println("p2.slopeTo(p5) is "+p2.slopeTo(p5));
        }
    }
    
    

    二:BruteCollinearPoints.java

    import java.util.ArrayList;
    import java.util.Arrays;
    import edu.princeton.cs.algs4.In;
    import edu.princeton.cs.algs4.Insertion;
    import edu.princeton.cs.algs4.StdOut;
    import edu.princeton.cs.algs4.StdDraw;
    
    public class BruteCollinearPoints {
         //to store line segments
         private ArrayList<LineSegment> LineSegmentList;
         //to store the given points
         private Point[] points;
        
         //在构造函数中找出由4点构成的线段(作业说了:没有5点及更多点共线的情况)
         // finds all line segments containing 4 point
         public BruteCollinearPoints(Point[] pointsIn) { 
             //三种异常
             //一:if the argument to the constructor is null
             System.out.println(" a ");
             if(pointsIn == null) throw new IllegalArgumentException("there is no point");
             //二:if any point in the array is null
             int N = pointsIn.length;
             for(int i=0;i<N;i++) if(pointsIn[i]==null) throw new IllegalArgumentException("exist null point");
             //三:if the argument to the constructor contains a repeated point
             //检查是否有重复的点,先排序,再查重会方便,作业允许这样: For example, you may use Arrays.sort()
             points = new Point[N];
             for(int i=0;i<N;i++) points[i] = pointsIn[i];
             Arrays.sort(points);
             for(int i=1;i<N;i++) if(points[i-1].compareTo(points[i])==0) throw new IllegalArgumentException("exist repeated point"); 
             
             //to save every required line segment
             LineSegmentList = new ArrayList<LineSegment>();
                 
             //find line segment
             for(int dot1=0;dot1<=N-4;dot1++) {
                 for(int dot2=dot1+1;dot2<=N-3;dot2++) {
                     //k12:the slope between point[dot2] and point[dot1]
                     double k12 = points[dot2].slopeTo(points[dot1]);
                     for(int dot3=dot2+1;dot3<=N-2;dot3++) {
                         //k13:the slope between point[dot3] and point[dot1]
                         double k13 = points[dot3].slopeTo(points[dot1]);
                         if(k13 != k12) continue;
                         for(int dot4=dot3+1;dot4<=N-1;dot4++) {
                            //k14:the slope between point[dot4] and point[dot1]
                             double k14 = points[dot4].slopeTo(points[dot1]);
                             if(k14 != k12) continue;
                             //find a line segment
                             LineSegment linesegment = new LineSegment(points[dot1],points[dot4]);
                             LineSegmentList.add(linesegment);
                         }
                     }
                 }
             }
         }
         // the number of line segments
         public int numberOfSegments() {
             return LineSegmentList.size();
         }
         // the line segments
         public LineSegment[] segments() {
             LineSegment[] segments = new LineSegment[LineSegmentList.size()];
             int index=0;
             for(LineSegment Line : LineSegmentList) {
                 segments[index++] = Line;
             }
             return segments;
         }    
         //main
         public static void main(String[] args) {
                In in = new In("src/week3/input8.txt"); 
                int n = in.readInt();
                StdOut.println("total "+n+" points");
                Point[] points = new Point[n];
                for (int i = 0; i < n; i++) {
                    int x = in.readInt();
                    int y = in.readInt();
                    StdOut.println("("+x+","+y+")"); 
                    points[i] = new Point(x,y);
                }       
                //draw the points
                StdDraw.enableDoubleBuffering();
                StdDraw.setXscale(0, 32768);
                StdDraw.setYscale(0, 32768);
                StdDraw.setPenColor(StdDraw.RED);
                StdDraw.setPenRadius(0.01);
                for (Point p : points) {
                    p.draw();
                }
                StdDraw.show();              
                // print and draw the line segments
                BruteCollinearPoints collinear = new BruteCollinearPoints(points);
                StdOut.println(collinear.numberOfSegments());
                for (LineSegment segment : collinear.segments()) {
                    StdOut.println(segment);
                    segment.draw();
                }
                StdDraw.show();          
         }
    }
    

    三:FastCollinearPoints.java

    import java.util.ArrayList;
    import java.util.Arrays;
    import edu.princeton.cs.algs4.In;
    import edu.princeton.cs.algs4.StdDraw;
    import edu.princeton.cs.algs4.StdOut;
    
    //much faster than the brute-force solution
    public class FastCollinearPoints {
        //to store line segments
        private ArrayList<LineSegment> LineSegmentList;
        //to store the given points
        private Point[] points;
        
        // finds all line segments containing 4 or more points
        public FastCollinearPoints(Point[] pointsIn) {
             //三种异常
             //一:if the argument to the constructor is null
             if(pointsIn == null) throw new IllegalArgumentException("there is no point");
            //number of points
             int N=pointsIn.length;
             //二:if any point in the array is null
             for(int i=0;i<N;i++) if(pointsIn[i]==null) throw new IllegalArgumentException("exist null point");
             //三:if the argument to the constructor contains a repeated point
             //检查是否有重复的点,先排序,再查重会方便,作业允许这样: For example, you may use        Arrays.sort()
             //同时有利于后面排除重复线段
             points = new Point[N];
             for(int i=0;i<N;i++) points[i] = pointsIn[i];
             //用的是结合了归并和插入的tim排序,稳定排序
             Arrays.sort(points);
             for(int i=1;i<N;i++) if(points[i-1].compareTo(points[i])==0) throw new IllegalArgumentException("exist repeated point");
            
             //to save every required line segment
             LineSegmentList = new ArrayList<LineSegment>();
                 
             
             //当前的参考点
             Point currentCheck;
             //对照点重新存储,不包括参考点,共N-1个
             Point[] otherPoints = new Point[N-1];
             //开始比较斜率,一个一个来
             for (int i=0;i<N;i++) {
                 currentCheck = points[i];
                 // copy points without Point currentCheck to otherPoints
                 for(int j=0;j<N;j++) {
                     if(j<i) otherPoints[j] = points[j];
                     if(j>i) otherPoints[j-1] = points[j];
                 }
                 
                 //根据斜率对点排序
                 //用的是结合了归并和插入的tim排序,稳定排序
                 Arrays.sort(otherPoints,currentCheck.slopeOrder());
                 //遍历已经排序的otherPoints找线段
                 //注意,归并和插入排序都是稳定的,所以tim排序是稳定的,这非常重要
                 //配合Point的compareTo方法,可以直接过滤掉重复线段
                 //一开始没太注意compareTo方法,后来发现这个方法能固定住点之间的相对位置,所以可以过滤重复线段
                 //两点共线
                 int count=2;
                 for(int k=1;k<N-1;k++) {
                     double k1 = otherPoints[k-1].slopeTo(currentCheck);
                     double k2 = otherPoints[k].slopeTo(currentCheck);
                     if(k1==k2) {
                         count++;
                         //当循环到最后一个点,同时这个点和前面的点共线
                         if(k==N-2) {
                             //如果4点及以上共线,并且otherPoints中与参考点共线且排在最左边的点比参考点大的话,注意此处是遍历到头,所以索引是k-count+2
                             if(count>=4 && currentCheck.compareTo(otherPoints[k-count+2])==-1) { 
                                 //线段起点
                                 Point start = currentCheck;
                                 //线段终点
                                 Point end = otherPoints[k];
                                 LineSegment linesegment = new LineSegment(start,end);
                                 LineSegmentList.add(linesegment);
                             }
                         }
                     }
                     else{
                        //如果4点及以上共线,并且otherPoints中与参考点共线且排在最左边的点比参考点大的话,索引是k-count+1
                         if(count>=4 && currentCheck.compareTo(otherPoints[k-count+1])==-1) {
                                 Point start = currentCheck;
                                 Point end = otherPoints[k-1];
                                 LineSegment linesegment = new LineSegment(start,end);
                                 LineSegmentList.add(linesegment);
                         }
                         count=2;
                     }
                 }
             }
        }
    
        // the number of line segments
        public  int numberOfSegments() {
            return LineSegmentList.size();
        }
        // the line segments
        public LineSegment[] segments() {
            LineSegment[] segments = new LineSegment[LineSegmentList.size()];
             int index=0;
             for(LineSegment Line : LineSegmentList) {
                 segments[index++] = Line;
             }
             return segments;
        }
        
             //main
             public static void main(String[] args) {
                    In in = new In("src/week3/input9.txt"); 
                    int n = in.readInt();
                    StdOut.println("total "+n+" points");
                    Point[] points = new Point[n];
                    for (int i = 0; i < n; i++) {
                        int x = in.readInt();
                        int y = in.readInt();
                        StdOut.println("("+x+","+y+")"); 
                        points[i] = new Point(x,y);
                    }
                    
                    //draw the points
                    StdDraw.enableDoubleBuffering();
                    StdDraw.setXscale(0, 32768);
                    StdDraw.setYscale(0, 32768);
                    StdDraw.setPenColor(StdDraw.RED);
                    StdDraw.setPenRadius(0.01);
                    for (Point p : points) {
                        p.draw();
                    }
                    StdDraw.show();
                          
                    //print and draw the line segments
                    FastCollinearPoints collinear = new FastCollinearPoints(points);
                    StdOut.println(collinear.numberOfSegments());
                    for (LineSegment segment : collinear.segments()) {
                        StdOut.println(segment);
                        segment.draw();
                    }
                    StdDraw.show();          
             }
    }
    

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