算法导论第四版学习——习题三Collinear Points
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题目正文：
http://coursera.cs.princeton.edu/algs4/assignments/collinear.html
作业难点：
1、仔细思考会感觉有很多实现方法，但是如果没有适当使用排序，算法时间复杂度就会轻易超过要求。（包括暴力算法）
2、隐含需要实现自己的数据结构用来组织“线”的集合，当然也可以用Stack或者Queue或者LinkedList，但是我个人是自己实现的。
3、由于一条线段只能出现一次，所以有一个去重的问题。（最难）
作业技巧：
1、基本按照题目的先后顺序实现，结合视频学习中的一些知识点就可以把任务串起来，不会无从下手。
2、SlopTo的计算结果和点数组的排列顺序有关吗？A.SlopeTo(B)和B.SlopeTo(A)其实是一样的，合理利用cache防止多次计算。
3、在暴力方法中，引入排序可以解决去重问题。
4、如果去重是通过在结果集合中查找是否存在，就铁定会超过复杂度要求，所以必须在添加结果集之前就能判断。
这里就有一个重要推论：a-b-c-d线段，无论从a开始查找还是从b开始查找，线段最大和最小点都不变为a和d，所以我们只需要找到起始点为线段最小点的线段就可以了，其他线段均抛弃即可
代码参考：
（这是我自己亲测100分的答案，不代表写得最好，请在自己实在完成不了的时候再看，不然的话做这个题目的意义一点都没有）
1 import edu.princeton.cs.algs4.StdDraw; 2 3 /****************************************************************************** 4 * Compilation: javac Point.java 5 * Execution: java Point 6 * Dependencies: none 7 * 8 * An immutable data type for points in the plane. 9 * For use on Coursera, Algorithms Part I programming assignment. 10 * 11 ******************************************************************************/ 12 import java.util.Comparator; 13 14 15 public class Point implements Comparable<Point> { 16 private final int x; // x-coordinate of this point 17 private final int y; // y-coordinate of this point 18 19 /** 20 * Initializes a new point. 21 * 22 * @param x the <em>x</em>-coordinate of the point 23 * @param y the <em>y</em>-coordinate of the point 24 */ 25 public Point(int x, int y) { 26 /* DO NOT MODIFY */ 27 this.x = x; 28 this.y = y; 29 } 30 31 /** 32 * Draws this point to standard draw. 33 */ 34 public void draw() { 35 /* DO NOT MODIFY */ 36 StdDraw.point(x, y); 37 } 38 39 /** 40 * Draws the line segment between this point and the specified point 41 * to standard draw. 42 * 43 * @param that the other point 44 */ 45 public void drawTo(Point that) { 46 /* DO NOT MODIFY */ 47 StdDraw.line(this.x, this.y, that.x, that.y); 48 } 49 50 /** 51 * Returns the slope between this point and the specified point. 52 * Formally, if the two points are (x0, y0) and (x1, y1), then the slope 53 * is (y1 - y0) / (x1 - x0). For completeness, the slope is defined to be 54 * +0.0 if the line segment connecting the two points is horizontal; 55 * Double.POSITIVE_INFINITY if the line segment is vertical; 56 * and Double.NEGATIVE_INFINITY if (x0, y0) and (x1, y1) are equal. 57 * 58 * @param that the other point 59 * @return the slope between this point and the specified point 60 */ 61 public double slopeTo(Point that) { 62 if ((this.x == that.x) && (this.y == that.y)) { 63 return Double.NEGATIVE_INFINITY; 64 } else if (this.x == that.x) { 65 return Double.POSITIVE_INFINITY; 66 } else if (this.y == that.y) { 67 return 0; 68 } else { 69 return (that.y - this.y) / (double) (that.x - this.x); 70 } 71 } 72 73 /** 74 * Compares two points by y-coordinate, breaking ties by x-coordinate. 75 * Formally, the invoking point (x0, y0) is less than the argument point 76 * (x1, y1) if and only if either y0 < y1 or if y0 = y1 and x0 < x1. 77 * 78 * @param that the other point 79 * @return the value <tt>0</tt> if this point is equal to the argument 80 * point (x0 = x1 and y0 = y1); 81 * a negative integer if this point is less than the argument 82 * point; and a positive integer if this point is greater than the 83 * argument point 84 */ 85 public int compareTo(Point that) { 86 if ((this.x == that.x) && (this.y == that.y)) { 87 return 0; 88 } else if ((that.y > this.y) || 89 ((that.y == this.y) && (that.x > this.x))) { 90 return -1; 91 } else { 92 return 1; 93 } 94 } 95 96 /** 97 * Compares two points by the slope they make with this point. 98 * The slope is defined as in the slopeTo() method. 99 * 100 * @return the Comparator that defines this ordering on points 101 */ 102 public Comparator<Point> slopeOrder() { 103 return new SlopeOrder(this); 104 } 105 106 /** 107 * Returns a string representation of this point. 108 * This method is provide for debugging; 109 * your program should not rely on the format of the string representation. 110 * 111 * @return a string representation of this point 112 */ 113 public String toString() { 114 /* DO NOT MODIFY */ 115 return "(" + x + ", " + y + ")"; 116 } 117 118 /** 119 * Unit tests the Point data type. 120 */ 121 public static void main(String[] args) { 122 /* YOUR CODE HERE */ 123 } 124 125 private class SlopeOrder implements Comparator<Point> { 126 private Point p0; 127 128 public SlopeOrder(Point invokePoint) { 129 p0 = invokePoint; 130 } 131 132 public int compare(Point p1, Point p2) { 133 double d01 = p0.slopeTo(p1); 134 double d02 = p0.slopeTo(p2); 135 136 if (d01 < d02) { 137 return -1; 138 } else if (d01 > d02) { 139 return 1; 140 } else { 141 return 0; 142 } 143 } 144 } 145 }
1 import edu.princeton.cs.algs4.In; 2 import edu.princeton.cs.algs4.StdDraw; 3 import edu.princeton.cs.algs4.StdOut; 4 import java.util.Arrays; 5 6 public class BruteCollinearPoints { 7 private int lineNumber; 8 private Node last; 9 10 public BruteCollinearPoints(Point[] points) // finds all line segments containing 4 points 11 { 12 13 if (points == null) { 14 throw new NullPointerException(); 15 } 16 17 lineNumber = 0; 18 19 int num = points.length; 20 21 Point[] clone = new Point[num]; 22 23 for (int i = 0; i < num; i++) { 24 if (points[i] == null) { 25 throw new NullPointerException(); 26 } 27 28 for (int j = i + 1; j < num; j++) { 29 if (points[i].compareTo(points[j]) == 0) { 30 throw new IllegalArgumentException(); 31 } 32 } 33 clone[i] = points[i]; 34 } 35 Arrays.sort(clone); 36 for (int i = 0; i < num; i++) { 37 for (int j = i+1; j < num; j++) { 38 for (int m = j+1; m < num; m++) { 39 for (int n = m+1; n < num; n++) { 40 double d01 = clone[i].slopeTo(clone[j]); 41 double d02 = clone[j].slopeTo(clone[m]); 42 double d03 = clone[m].slopeTo(clone[n]); 43 44 if (d01 == d02 && d02 == d03) { 45 if (last != null) { 46 Node newNode = new Node(); 47 newNode.prev = last; 48 newNode.value = new LineSegment(clone[i], 49 clone[n]); 50 last = newNode; 51 } else { 52 last = new Node(); 53 last.value = new LineSegment(clone[i], 54 clone[n]); 55 } 56 57 lineNumber++; 58 } 59 } 60 } 61 } 62 } 63 } 64 65 public int numberOfSegments() // the number of line segments 66 { 67 return lineNumber; 68 } 69 70 public LineSegment[] segments() // the line segments 71 { 72 LineSegment[] lines = new LineSegment[lineNumber]; 73 Node current = last; 74 75 for (int i = 0; i < lineNumber; i++) { 76 lines[i] = current.value; 77 current = current.prev; 78 } 79 80 return lines; 81 } 82 83 public static void main(String[] args) { 84 // read the n points from a file 85 In in = new In(args[0]); 86 int n = in.readInt(); 87 Point[] points = new Point[n]; 88 89 for (int i = 0; i < n; i++) { 90 int x = in.readInt(); 91 int y = in.readInt(); 92 points[i] = new Point(x, y); 93 } 94 95 // draw the points 96 StdDraw.enableDoubleBuffering(); 97 StdDraw.setXscale(0, 32768); 98 StdDraw.setYscale(0, 32768); 99 100 for (Point p : points) { 101 p.draw(); 102 } 103 104 StdDraw.show(); 105 106 // print and draw the line segments 107 BruteCollinearPoints collinear = new BruteCollinearPoints(points); 108 109 for (LineSegment segment : collinear.segments()) { 110 StdOut.println(segment); 111 segment.draw(); 112 } 113 114 StdDraw.show(); 115 } 116 117 private class Node { 118 private LineSegment value; 119 private Node prev; 120 } 121 }
1 import edu.princeton.cs.algs4.In; 2 import edu.princeton.cs.algs4.StdDraw; 3 import edu.princeton.cs.algs4.StdOut; 4 5 import java.util.Arrays; 6 7 public class FastCollinearPoints { 8 private int lineNumber; 9 private Node last; 10 11 public FastCollinearPoints(Point[] points) // finds all line segments containing 4 or more points 12 { 13 if (points == null) { 14 throw new NullPointerException(); 15 } 16 17 lineNumber = 0; 18 19 int num = points.length; 20 Point[] clone = new Point[num]; 21 22 for (int i = 0; i < num; i++) { 23 if (points[i] == null) { 24 throw new NullPointerException(); 25 } 26 27 for (int j = i + 1; j < num; j++) { 28 if (points[i].compareTo(points[j]) == 0) { 29 throw new IllegalArgumentException(); 30 } 31 } 32 clone[i] = points[i]; 33 } 34 Arrays.sort(clone); 35 36 if (num < 4) { 37 return; 38 } 39 40 for (int i = 0; i < num - 1; i++) { 41 int tempPointsNum = 0; 42 Point[] tempPoints = new Point[num - 1]; 43 44 for (int j = 0; j < num; j++) { 45 if (i != j) tempPoints[tempPointsNum++] = clone[j]; 46 } 47 48 Arrays.sort(tempPoints, clone[i].slopeOrder()); 49 50 int count = 0; 51 Point min = null; 52 Point max = null; 53 54 for (int j = 0; j < (tempPointsNum - 1); j++) { 55 if (clone[i].slopeTo(tempPoints[j]) == clone[i].slopeTo( 56 tempPoints[j + 1])) { 57 if (min == null) { 58 if (clone[i].compareTo(tempPoints[j]) > 0) { 59 max = clone[i]; 60 min = tempPoints[j]; 61 } else { 62 max = tempPoints[j]; 63 min = clone[i]; 64 } 65 } 66 67 if (min.compareTo(tempPoints[j + 1]) > 0) { 68 min = tempPoints[j + 1]; 69 } 70 71 if (max.compareTo(tempPoints[j + 1]) < 0) { 72 max = tempPoints[j + 1]; 73 } 74 75 count++; 76 77 if (j == (tempPointsNum - 2)) { 78 if (count >= 2 && clone[i].compareTo(min) == 0) { 79 addLine(min, max); 80 } 81 82 count = 0; 83 min = null; 84 max = null; 85 } 86 } else { 87 if (count >= 2 && clone[i].compareTo(min) == 0) { 88 addLine(min, max); 89 } 90 91 count = 0; 92 min = null; 93 max = null; 94 } 95 } 96 } 97 } 98 99 private void addLine(Point a, Point b) { 100 if (last != null) { 101 Node newNode = new Node(); 102 newNode.prev = last; 103 newNode.value = new LineSegment(a, b); 104 last = newNode; 105 } else { 106 last = new Node(); 107 last.value = new LineSegment(a, b); 108 } 109 lineNumber++; 110 } 111 112 public int numberOfSegments() // the number of line segments 113 { 114 return lineNumber; 115 } 116 117 public LineSegment[] segments() // the line segments 118 { 119 LineSegment[] lines = new LineSegment[lineNumber]; 120 Node current = last; 121 122 for (int i = 0; i < lineNumber; i++) { 123 lines[i] = current.value; 124 current = current.prev; 125 } 126 127 return lines; 128 } 129 130 public static void main(String[] args) { 131 // read the n points from a file 132 In in = new In(args[0]); 133 int n = in.readInt(); 134 Point[] points = new Point[n]; 135 136 for (int i = 0; i < n; i++) { 137 int x = in.readInt(); 138以上是关于算法导论第四版学习——习题三Collinear Points的主要内容，如果未能解决你的问题，请参考以下文章