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总结
(代码有详细注释)
- 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
- 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)
- 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]) - 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
- 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
-
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 {
// 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 slopeOrder(){
return new SlopeOrder();
}
//nested class
//sort rule
private class SlopeOrder implements Comparator{
public int compare(Point p,Point q) {
//p点斜率大
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 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();
//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 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();
//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;ii) 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=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();
}
}