《Java语言程序设计(基础篇)》(第10版 梁勇 著)第二十二章练习题答案

《Java语言程序设计（基础篇）》（第10版梁勇著）第二十二章练习题答案22.1public class Exercise22_01 {public static void main(String[] args) {java.util.Scanner input = new java.util.Scanner(System.in);System.out.print("Enter a string: ");String s = input.nextLine();System.out.println("Maximum consecutive substring is " + maxConsecutiveSortedSubstring(s));}/*** The worst-case complexity is O(n^2), where n is s.length(). Can you improve * the algorithm?*/public static String maxConsecutiveSortedSubstring(String s) {int[] maxConsecutiveLength = new int[s.length()];int current = 0;for (int i = 1; i < s.length(); i++) {if (s.charAt(i) <= s.charAt(i - 1)) { // next character is smallercurrent = i;} else {// s[current], s[current+1], ..., s[i-1] is a maximal increasing// sequencefor (int j = i - 1; j >= current; j--)maxConsecutiveLength[j]++;}}// maxConsecutiveLength[i] denotes the length of a max sequence starting at// index iint currentMaxLength = maxConsecutiveLength[0];int index = 0;for (int i = 0; i < s.length(); i++) {if (maxConsecutiveLength[i] > currentMaxLength) {currentMaxLength = maxConsecutiveLength[i];index = i;}}return s.substring(index, index + currentMaxLength + 1);}/* O(n) version, 9/21/2009 */public static String maxConsecutiveSortedSubstring1(String s) {int currentMaxLength = 1;int lastIndexOfMaxConsecutiveSortedSubstring = 0;int possibleMaxLength = 1;for (int i = 1; i < s.length(); i++) {if (s.charAt(i) > s.charAt(i - 1)) {if (lastIndexOfMaxConsecutiveSortedSubstring == i - 1) {// Add s(i) into the max consecutive substringcurrentMaxLength++;lastIndexOfMaxConsecutiveSortedSubstring++;} else {possibleMaxLength++;if (possibleMaxLength > currentMaxLength) {currentMaxLength = possibleMaxLength;lastIndexOfMaxConsecutiveSortedSubstring = i;possibleMaxLength = 1;}}}}return s.substring(lastIndexOfMaxConsecutiveSortedSubstring- currentMaxLength + 1, lastIndexOfMaxConsecutiveSortedSubstring + 1); }}22.1附加import javafx.application.Application;import javafx.geometry.Pos;import javafx.stage.Stage;import javafx.scene.Scene;import bel;import javafx.scene.image.Image;import javafx.scene.image.ImageView;import yout.BorderPane;import yout.Pane;import javafx.scene.paint.Color;import javafx.scene.shape.Line;import javafx.scene.shape.Rectangle;public class Exercise22_01Extra extends Application {public static final int SIZE = 8; // The size of the chess board// queens are placed at (i, queens[i])// -1 indicates that no queen is currently placed in the ith row// Initially, place a queen at (0, 0) in the 0th rowprivate int[] queens = {-1, -1, -1, -1, -1, -1, -1, -1};private Label lblStatus = new Label();// Display chess boardChessBoard chessBoard = new ChessBoard();int k = 0;@Override // Override the start method in the Application classpublic void start(Stage primaryStage) {BorderPane pane = new BorderPane();Label lblMessage = new Label("Click the mouse to see the next move"); pane.setBottom(lblMessage);BorderPane.setAlignment(lblMessage, Pos.CENTER);pane.setTop(lblStatus);BorderPane.setAlignment(lblStatus, Pos.CENTER);pane.setCenter(chessBoard);// Create a scene and place it in the stageScene scene = new Scene(pane, 50 * SIZE, 55 * SIZE);primaryStage.setTitle("Exercise22_01"); // Set the stage titleprimaryStage.setScene(scene); // Place the scene in the stageprimaryStage.show(); // Display the stagechessBoard.setOnMouseClicked(e -> {search();if (k == -1) {lblStatus.setText("No solution found");chessBoard.setOnMouseClicked(null);}else if (k == 8) {lblStatus.setText("A solution found");chessBoard.setOnMouseClicked(null);}chessBoard.repaint();});chessBoard.repaint();}/** Search for a solution */private boolean search() {// k - 1 indicates the number of queens placed so far// We are looking for a position in the kth row to place a queen if (k >= 0 && k < SIZE) {// Find a position to place a queen in the kth rowint j = findPosition(k);if (j < 0) {queens[k] = -1;k--; // back track to the previous row} else {queens[k] = j;k++;}}if (k == -1)return false; // No solutionelsereturn true; // A solution is found}public int findPosition(int k) {int start = queens[k] + 1; // Search for a new placementfor (int j = start; j < SIZE; j++) {if (isValid(k, j))return j; // (k, j) is the place to put the queen now}return -1;}/** Return true if a queen can be placed at (row, column) */ public boolean isValid(int row, int column) {for (int i = 1; i <= row; i++)if (queens[row - i] == column // Check column|| queens[row - i] == column - i // Check upleft diagonal|| queens[row - i] == column + i) // Check upright diagonal return false; // There is a conflictreturn true; // No conflict}class ChessBoard extends Pane {private Image queenImage = new Image("image/queen.jpg");ChessBoard() {setStyle("-fx-border-color: black");}double w = 400;double h = 400;public void repaint() {getChildren().clear();// Highlight the current rowRectangle rectangle = new Rectangle(0, k * h / SIZE,w, h / SIZE);rectangle.setFill(Color.YELLOW);getChildren().add(rectangle);// Paint the queensfor (int i = 0; i < SIZE; i++) {int j = queens[i]; // The position of the queen in row i System.out.println(j);System.out.println("getWidth " + w);ImageView imageView = new ImageView(queenImage);imageView.setX(j * w / SIZE);imageView.setY(i * h / SIZE);imageView.setFitWidth(w / SIZE);imageView.setFitHeight(h / SIZE);getChildren().add(imageView);}// Draw the horizontal and vertical linesfor (int i = 1; i < SIZE; i++) {getChildren().addAll(new Line(0, i * h / SIZE,w, i * h / SIZE), new Line(i * w / SIZE, 0,i * w / SIZE, h));}}}/*** The main method is only needed for the IDE with limited* JavaFX support. Not needed for running from the command line.*/public static void main(String[] args) {launch(args);}}22.2public class Exercise22_02 {public static void main(String[] args) {java.util.Scanner input = new java.util.Scanner(System.in);System.out.print("Enter a string: ");String s = input.nextLine();System.out.println("Maximum increasing ordered subsequence is " + maxSubstring(s));}/*** The worst-case complexity is O(n^2)*/public static String maxSubstring(String s) {// maxLength[i] stores the length of the max substring ending at index i int[] maxLength = new int[s.length()];// previous[i] stores the index of the previous element in the sequenece int[] previous = new int[s.length()];for (int i = 0; i < s.length(); i++) {previous[i] = -1;for (int j = i - 1; j >= 0; j--) {if (s.charAt(i) > s.charAt(j) &&maxLength[i] < maxLength[j] + 1) {maxLength[i] = maxLength[j] + 1;previous[i] = j;}}}// Find the largest subsequence length and ending indexint maxL = maxLength[0];int index = 0;for (int i = 1; i < s.length(); i++) {if (maxL < maxLength[i]) {maxL = maxLength[i];index = i;}}// Construct the subsequence by tracing through previous char[] chars = new char[maxL + 1];int i = maxL;while (index != -1) {chars[i--] = s.charAt(index);index = previous[index];}return new String(chars);}}22.3public class Exercise22_03 {public static void main(String[] args) {java.util.Scanner input = new java.util.Scanner(System.in);System.out.print("Enter a string s1: ");String s1 = input.nextLine();System.out.print("Enter a string s2: ");String s2 = input.nextLine();int index = match1(s1, s2);if (index >= 0)System.out.println("matched at index " + index);elseSystem.out.println("unmatched");}/*** The worst-case complexity is O(n), where n is s.length() ** @param s String* @param pattern String* @return int*/public static int match(String s, String pattern) {int k = 0;for (int i = 0; i < s.length(); i++) {if (k == pattern.length())return i - pattern.length();else {if (s.charAt(i) == pattern.charAt(k))k++;elsek = 0;}}if (k == pattern.length())return s.length() - pattern.length();elsereturn -1;}/*** This is a brute-force approach, O(|s|*|pattern|))*/public static int match1(String s, String pattern) {for (int i = 0; i < s.length(); i++) {// Check if pattern matches si, si+1, ...int k = 0;for ( ; k < pattern.length(); k++) {if (s.charAt(i + k) != pattern.charAt(k))break;}if (k == pattern.length())return i;}return -1;}}22.4public class Exercise22_04 {public static void main(String[] args) {java.util.Scanner input = new java.util.Scanner(System.in); System.out.print("Enter a string s1: ");String s1 = input.nextLine();System.out.print("Enter a string s2: ");String s2 = input.nextLine();int index = match(s1, s2);if (index >= 0)System.out.println("matched at index " + index);elseSystem.out.println("unmatched");// System.out.println(match("aerbbbcbbb", "bbcb"));}/*** The worst-case complexity is O(nk), where n is s.length() and k * is pattern.length()** @param s String* @param pattern String* @return int*/public static int match(String s, String pattern) {for (int p = 0; p < s.length(); p++) {int k = 0;for (int i = p; i < s.length(); i++) {if (k == pattern.length())return i - pattern.length();else {if (s.charAt(i) == pattern.charAt(k))k++;elsebreak;}}if (k == pattern.length())return s.length() - pattern.length();}return -1;}}22.5import java.util.Scanner;public class Exercise22_05 {public static void main(String[] args) {Scanner input = new Scanner(System.in);System.out.print("Enter a series of numbers ending with 0: ");int longestSequenceCount = 0;int longestSequenceValue = 0;int longestSequenceIndex = 0;int currentLongestSequenceCount = 1;int currentLongestSequenceValue = 0;int value;int index = 0;int previous = 0;do {value = input.nextInt();if (value == previous)currentLongestSequenceCount++;else if (currentLongestSequenceCount > longestSequenceCount) {longestSequenceCount = currentLongestSequenceCount;longestSequenceValue = currentLongestSequenceValue;longestSequenceIndex = index - currentLongestSequenceCount;currentLongestSequenceCount = 1; // Reset the countcurrentLongestSequenceValue = value; // Reset the value}previous = value;index++;}while (value != 0);System.out.println("The longest same number sequence starts at index " + longestSequenceIndex + " with " + longestSequenceCount+ " values of " + longestSequenceValue);}}22.6public class Exercise22_06 {public static void main(String[] args) {// Find the first 47 Fibonacci numbersfinal int INDEX = 47;int[] numbers = new int[INDEX];numbers[0] = 0;numbers[1] = 1;for (int i = 2; i < INDEX; i++) {numbers[i] = numbers[i - 1] + numbers[i - 2];}System.out.println("\t\t\t40\t41\t42\t43\t44\t45");System.out.println("-----------------------------------------------");System.out.print("Listing 23.2 GCD1");long[] executionTime = new long[6];for (int i = 40; i <= 45; i++) {long startTime = System.currentTimeMillis();gcd1(numbers[i], numbers[i + 1]);executionTime[i - 40] = System.currentTimeMillis() - startTime;}for (int i = 0; i <= 5; i++) {System.out.print("\t" + executionTime[i]);}System.out.print("\nListing 23.3 GCD2");for (int i = 40; i <= 45; i++) {long startTime = System.currentTimeMillis();gcd2(numbers[i], numbers[i + 1]);executionTime[i - 40] = System.currentTimeMillis() - startTime;}for (int i = 0; i <= 5; i++) {System.out.print("\t" + executionTime[i]);}}/** From Listing 21.2: Find gcd for intergers m and n */public static int gcd1(int m, int n) {int gcd = 1;if (m == n) return m;for (int k = 1; k <= m / 2 && k <= n / 2; k++) {if (m % k == 0 && n % k == 0)gcd = k;}return gcd;}/** From Listing 21.3: Find gcd for intergers m and n */public static int gcd2(int m, int n) {if (m % n == 0)return n;elsereturn gcd2(n, m % n);}}22.7import java.util.Arrays;import java.util.ArrayList;/** This program works, but the design is not good. Redesign it */ public class Exercise22_07 {public static void main(String[] args) {double[][] points = new double[1000][2];for (int i = 0; i < points.length; i++) {points[i][0] = Math.random() * 100;points[i][1] = Math.random() * 100;}long startTime = System.currentTimeMillis();Pair pair = closestPairBruteForce(points);System.out.println("shortest distance is " + pair.getDistance()); System.out.println(pair);long endTime = System.currentTimeMillis();System.out.println("Time spent on the brute-force algorithm is " + (endTime - startTime) + " milliseconds");startTime = System.currentTimeMillis();Pair closestPair = getClosestPair(points);System.out.println("shortest distance is " + closestPair.getDistance()); System.out.println(closestPair);endTime = System.currentTimeMillis();System.out.println("Time spent on the divide-and-conquer algorithm is " + (endTime - startTime) + " milliseconds");}/** Return the distance of the closest pair of points */public static Pair getClosestPair(double[][] points) {Point[] pointsOrderedOnX = new Point[points.length];for (int i = 0; i < pointsOrderedOnX.length; i++)pointsOrderedOnX[i] = new Point(points[i][0], points[i][1]);return getClosestPair(pointsOrderedOnX);}/** Return the distance of the closest pair of points */public static Pair getClosestPair(Point[] points) {Arrays.sort(points);// Locate the identical points if existsPair pair = checkIdentical(points);if (pair != null)return pair; // The distance between the identical points is 0Point[] pointsOrderedOnY = points.clone();Arrays.sort(pointsOrderedOnY, new CompareY());return distance(points, 0, points.length - 1, pointsOrderedOnY);}/** Locate the identical points if exist */public static Pair checkIdentical(Point[] pointsOrderedOnX) {Pair pair = new Pair();for (int i = 0; i < pointsOrderedOnX.length - 1; i++) {if (pointsOrderedOnX[i].compareTo(pointsOrderedOnX[i + 1]) == 0) {pair.p1 = pointsOrderedOnX[i];pair.p2 = pointsOrderedOnX[i + 1];return pair;}}return null;}/** Return the distance of the closest pair of points* in pointsOrderedOnX[low..high]. This is a recursive method.* pointsOrderedOnX and pointsOrderedOnY are not changed in the* subsequent recursive calls.*/public static Pair distance(Point[] pointsOrderedOnX,int low, int high, Point[] pointsOrderedOnY) {if (low >= high) // Zero or one point in the setreturn null;else if (low + 1 == high) {// Two points in the setPair pair = new Pair();pair.p1 = pointsOrderedOnX[low];pair.p2 = pointsOrderedOnX[high];return pair;}// Step 2int mid = (low + high) / 2;Pair pair1 = distance(pointsOrderedOnX, low, mid, pointsOrderedOnY);Pair pair2 = distance(pointsOrderedOnX, mid + 1, high, pointsOrderedOnY);double d;Pair pair = null;if (pair1 == null && pair2 == null) {d = Double.MAX_VALUE;} else if (pair1 == null) {d = pair2.getDistance();pair = pair2;} else if (pair2 == null) {d = pair1.getDistance();pair = pair1;} else {d = Math.min(pair1.getDistance(), pair2.getDistance());if (pair1.getDistance() <= pair2.getDistance())pair = pair1;elsepair = pair2;}// Obtain stripL and stripRArrayList<Point> stripL = new ArrayList<Point>();ArrayList<Point> stripR = new ArrayList<Point>();for (int i = 0; i < pointsOrderedOnY.length; i++)if (pointsOrderedOnY[i].x <= pointsOrderedOnX[mid].x&& pointsOrderedOnY[i].x >= pointsOrderedOnX[mid].x - d)stripL.add(pointsOrderedOnY[i]);elsestripR.add(pointsOrderedOnY[i]);// Step 3: Find the closest pair for a point in stripL and// a point in stripRdouble d3 = d;int r = 0;for (int i = 0; i < stripL.size(); i++) {while (r < stripR.size() && stripL.get(i).y > stripR.get(r).y + d)r++;// Compare a point in stripL with at most six points in stripRint r1 = r; // Start from r1 up in stripRwhile (r1 < stripR.size() && stripR.get(r1).y <= stripL.get(i).y + d) {if (d3 > distance(stripL.get(i), stripR.get(r1))) {d3 = distance(stripL.get(i), stripR.get(r1));pair.p1 = stripL.get(i);pair.p2 = stripR.get(r1);}r1++;}}return pair;}/** Compute the distance between two points p1 and p2 */public static double distance(Point p1, Point p2) {return distance(p1.x, p1.y, p2.x, p2.y);}/** Compute the distance between points (x1, y1) and (x2, y2) */public static double distance(double x1, double y1,double x2, double y2) {return Math.sqrt((x2 - x1) * (x2 - x1) + (y2 - y1) * (y2 - y1));}/** Define a class for a point with x- and y- coordinates */static class Point implements Comparable<Point> {double x;double y;Point(double x, double y) {this.x = x;this.y = y;}@Overridepublic int compareTo(Point p2) {if (this.x < p2.x)return -1;else if (this.x == p2.x) {// Secondary order on y-coordinatesif (this.y < p2.y)return -1;else if (this.y == p2.y)return 0;elsereturn 1;} elsereturn 1;}}/*** A comparator for comparing points on their y-coordinates. If y-coordinates * are the same, compare their x-coordinator.*/static class CompareY implements parator<Point> {public int compare(Point p1, Point p2) {if (p1.y < p2.y)return -1;else if (p1.y == p2.y) {// Secondary order on x-coordinatesif (p1.x < p2.x)return -1;else if (p1.x == p2.x)return 0;elsereturn 1;} elsereturn 1;}}/** A class to represent two points */public static class Pair {Point p1;Point p2;public double getDistance() {/* if (p1 == null || p2 == null)return Double.MAX_VALUE;else */return distance(p1, p2);}@Overridepublic String toString() {return"(" + p1.x + ", " + p1.y + ") and (" + p2.x + ", " + p2.y + ")"; }}public static Pair closestPairBruteForce(double[][] points) {Point[] p = new Point[points.length];for (int i = 0; i < p.length; i++)p[i] = new Point(points[i][0], points[i][1]);return closestPairBruteForce(p);}public static Pair closestPairBruteForce(Point[] points) {Pair pair = new Pair();if (points.length < 2)return null;// p1 and p2 are the indices in the points arraypair.p1 = points[0];pair.p2 = points[1];double shortestDistances = distance(pair.p1, pair.p2); // Initialize shortestDistances// Compute distance for every two pointsfor (int i = 0; i < points.length; i++) {for (int j = i + 1; j < points.length; j++) {double distance = distance(points[i], points[j]); // Find distanceif (distance < shortestDistances) {pair.p1 = points[i];pair.p2 = points[j];shortestDistances = distance; // Update shortestDistances}}}return pair;}}22.8import java.io.*;public class Exercise22_08 {final static int ARRAY_SIZE = 100000;public static void main(String[] args) throws Exception {// A list to hold prime numbersfinal long N = 100000002; // Find primes up to Nlong[] primeNumbers = new long[ARRAY_SIZE];long number; // A number to be tested for primenessRandomAccessFile inout =new RandomAccessFile("PrimeNumbers.dat", "rw");if (inout.length() == 0) {number = 1;}else {inout.seek(inout.length() - 8); // A long is 8 bytesnumber = inout.readLong(); // Get the last prime number in the file }int squareRoot = 1;// Repeatedly find prime numbersnewNumber:while (number <= N) {// Check if 2, 3, 4, ..., N is primenumber++;inout.seek(0);if (squareRoot * squareRoot < number) {squareRoot++;}while (inout.getFilePointer() < inout.length()) {int size = readNextBatch(primeNumbers, inout);// Exercise03_21 if number is primefor (int k = 0; k < size && primeNumbers[k] <= squareRoot; k++) {if (number % primeNumbers[k] == 0) { // If true, not primecontinue newNumber; // Exit the for loop}}}// Append a new prime number to the end of the fileinout.seek(inout.length());inout.writeLong(number);}inout.close();}public static int readNextBatch(long[] primeNumbers, RandomAccessFile inout) {int size = 0;try {while (inout.getFilePointer() < inout.length() && size < ARRAY_SIZE) { primeNumbers[size++] = inout.readLong();}}catch (Exception ex) {ex.printStackTrace();}return size;}}22.9import java.util.ArrayList;import java.util.Scanner;public class Exercise22_09 {public static void main(String[] args) {Scanner input = new Scanner(System.in);System.out.print("How many points are in the set? ");int numberOfPoints = input.nextInt();double[][] p = new double[numberOfPoints][2];System.out.print("Enter " + p.length + " points: ");for (int i = 0; i < p.length; i++)for (int j = 0; j < p[i].length; j++)p[i][j] = input.nextDouble();ArrayList<MyPoint> list = getConvexHull(p);System.out.print("The convex hull is ");for (int i = 0; i < list.size(); i++) {System.out.print("(" + list.get(i).x + ", " + list.get(i).y + ") "); }}static class MyPoint {double x, y;MyPoint(double x, double y) {this.x = x; this.y = y;}public boolean equals(Object o) {return x == ((MyPoint)o).x && y == ((MyPoint)o).y;}}/** Return the points that form a convex hull */public static ArrayList<MyPoint> getConvexHull(double[][] s) {// For efficiency, create an array of objectsMyPoint[] myPoints = new MyPoint[s.length];for (int i = 0; i < myPoints.length; i++)myPoints[i] = new MyPoint(s[i][0], s[i][1]);// Step 1MyPoint h0 = getRightmostLowestPoint(myPoints);ArrayList<MyPoint> H = new ArrayList<MyPoint>();H.add(h0);MyPoint t0 = h0;// Step 2 and Step 3while (true) {// Step 2MyPoint t1 = myPoints[0]; // Let t1 be s0for (int i = 1; i < myPoints.length; i++) {double status = whichSide(t0.x, t0.y, t1.x, t1.y, myPoints[i].x, myPoints[i].y);if (status > 0) // Right side of the line. Please note we are using the Java coordinate system. y increases downwardt1 = myPoints[i];else if (status == 0) {if (distance(s[i][0], s[i][1], t0.x, t0.y) > distance(t1.x, t1.y, t0.x, t0.y))t1 = myPoints[i];}}// Step 3if (t1.equals(h0)) // (t1.x == h0.x && t1.y == h0.y)break; // A convex hull is foundelse {H.add(t1);t0 = t1;}}return H;}/** Return the rightmost lowest point in S */private static MyPoint getRightmostLowestPoint(MyPoint[] p) {int rightMostIndex = 0;double rightMostX = p[0].x;double rightMostY = p[0].y;for (int i = 1; i < p.length; i++) {if (rightMostY < p[i].y) {rightMostY = p[i].y;rightMostX = p[i].x;rightMostIndex = i;}else if (rightMostY == p[i].y && rightMostX < p[i].x) {rightMostX = p[i].x;rightMostIndex = i;}}return p[rightMostIndex];}public static double distance(double x1, double y1, double x2, double y2) { return Math.sqrt((x1 - x2) * (x1 - x2) + (y1 - y2) * (y1 - y2));}static double whichSide(double x0, double y0, double x1, double y1, double x2, double y2) {return (x1 - x0) * (y2 - y0) - (x2 - x0) * (y1 - y0);}}22.10import java.io.*;public class Exercise22_10 {public static void main(String[] args) throws Exception {int count = 0; // Count the number of prime numbersDataInputStream input =new DataInputStream(new BufferedInputStream(new FileInputStream("Exercise22_07.dat")));long[] limits = {10,100,1000,10000,100000,1000000,10000000,100000000,1000000000,10000000000L,100000000000L,1000000000000L,10000000000000L,100000000000000L,1000000000000000L,10000000000000000L,100000000000000000L,1000000000000000000L};int k = 0;while (input.available() > 0) {if (input.readLong() > limits[k]) {System.out.println("Number of prime number <= " + limits[k] + " is " + count);k++;if (k == limits.length) break;}count++;}input.close();System.out.println("Total number of prime numbers found: " +count);}}22.11import java.util.ArrayList;import java.util.Collections;import java.util.Scanner;public class Exercise22_11 {public static void main(String[] args) {Scanner input = new Scanner(System.in);System.out.print("How many points are in the set? ");int numberOfPoints = input.nextInt();double[][] p = new double[numberOfPoints][2];System.out.print("Enter " + p.length + " points: ");for (int i = 0; i < p.length; i++)for (int j = 0; j < p[i].length; j++)p[i][j] = input.nextDouble();ArrayList<MyPoint> list = getConvexHull(p);System.out.print("The convex hull is ");for (int i = 0; i < list.size(); i++) {System.out.print("(" + list.get(i).x + ", " + list.get(i).y + ") "); }}/** Return the points that form a convex hull */public static ArrayList<MyPoint> getConvexHull(double[][] points) {// Create MyPoint objects for pointsArrayList<MyPoint> p = new ArrayList<MyPoint>();for (int i = 0; i < points.length; i++)p.add(new MyPoint(points[i][0], points[i][1]));placeP0(p);for (int i = 0; i < points.length; i++)p.get(i).setRightMostLowestPoint(p.get(0));Collections.sort(p);p = discardTies(p); // If two points have the same angle, discard the one that is closer to p0if (p.size() < 3)return null;ArrayList<MyPoint> stack = new ArrayList<MyPoint>(); // We use ArrayList rather than java.util.Stack, because the algorithm needs to access the top two elementsstack.add(p.get(0));stack.add(p.get(1));stack.add(p.get(2));int i = 3;while (i < p.size()) {MyPoint t2 = stack.get(stack.size() - 1);MyPoint t1 = stack.get(stack.size() - 2);if(whichSide(t1.x, t1.y, t2.x, t2.y, p.get(i).x, p.get(i).y) >= 0) { // on the right of the line from t1 to t2// pop the top element off the stack。