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java - 快速排序算法的动画

转载 作者:行者123 更新时间:2023-12-01 09:58:55 24 4
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我正在尝试使用 JavaFX 库对快速排序算法的分区部分进行动画处理。我把程序的所有图形部分都写下来了,但我的步骤方法遇到了问题。我的步骤方法中有一个逻辑错误,我似乎无法弄清楚它是什么。每次单击步骤按钮时,步骤方法应该执行一步。

private void quickSort(int lowerIndex, int higherIndex) {

int i = lowerIndex;
int j = higherIndex;
// calculate pivot number, I am taking pivot as middle index number
int pivot = array[lowerIndex+(higherIndex-lowerIndex)/2];
// Divide into two arrays
while (i <= j) {
/**
* In each iteration, we will identify a number from left side which
* is greater then the pivot value, and also we will identify a number
* from right side which is less then the pivot value. Once the search
* is done, then we exchange both numbers.
*/
while (array[i] < pivot) {
i++;
}
while (array[j] > pivot) {
j--;
}
if (i <= j) {
exchangeNumbers(i, j);
//move index to next position on both sides
i++;
j--;
}
}
// call quickSort() method recursively
if (lowerIndex < j)
quickSort(lowerIndex, j);
if (i < higherIndex)
quickSort(i, higherIndex);
}

上面的代码是快速排序算法的源代码,step方法应该只执行快速排序算法的一步。我尝试在步骤方法中实现相同的逻辑,但我的逻辑是错误的。该程序的正确输出位于此链接 http://www.cs.armstrong.edu/liang/animation/web/QuickSortPartition.html以供引用。

下面是我正在使用的代码

 import java.util.Arrays;
import javafx.application.Application;
import javafx.geometry.Pos;
import javafx.scene.Scene;
import javafx.scene.control.Button;
import javafx.scene.layout.BorderPane;
import javafx.scene.layout.HBox;
import javafx.scene.layout.Pane;
import javafx.scene.paint.Color;
import javafx.scene.shape.Line;
import javafx.scene.shape.Rectangle;
import javafx.scene.text.Text;
import javafx.stage.Stage;

public class AnimateQuickSort extends Application {
public final static int ARRAY_SIZE = 20;
private int[] array = new int[ARRAY_SIZE];
private int leftP = 1;
private int rightP = array.length - 1;
private int pivot = 0;

public void start(Stage primaryStage) {
AnimationPane pane = new AnimationPane();

Button btStep = new Button("Step");
Button btReset = new Button("Reset");

HBox hBox = new HBox(5);
hBox.getChildren().addAll(btStep, btReset);
hBox.setAlignment(Pos.CENTER);

BorderPane borderPane = new BorderPane();
borderPane.setCenter(pane);
borderPane.setBottom(hBox);

// Create a scene and place it in the stage
Scene scene = new Scene(borderPane, 700, 225);
primaryStage.setTitle("QuickSort");
primaryStage.setScene(scene);
primaryStage.show();

initializeArray();

pane.repaint();

btStep.setOnAction(e -> {
if (step()) {
pane.repaint();
} else {
pane.repaint();
}
});

btReset.setOnAction(e -> {
reset();
pane.repaint();
});
}

public static void main(String[] args) {
launch(args);
}

public void initializeArray() {

for (int i = 0; i < array.length; i++) {
array[i] = (int) (Math.random() * 999 + 1);
}

// Arrays.sort(array);

}

public void reset() {
leftP = 1;
rightP = array.length - 1;
pivot = 0;
initializeArray();
}

public boolean step() {
// quicksort(array);
int begin = 0;
int pivotVal = array[0];
int array1;

if (leftP <= rightP && array[leftP] <= pivotVal) {
leftP++;
array1 = array[leftP];
array[leftP] = array[rightP];
array[rightP] = array1;
}

if (rightP >= leftP && array[rightP] > pivotVal) {
rightP--;
}
if (rightP > leftP) {
if (rightP != begin) {
array1 = array[rightP];
array[rightP] = array[begin];
array[begin] = array1;
}
}
// Return the index of the pivot element.
return true;
}

class AnimationPane extends Pane {
private int startingX = 20;
private int startingY = 20;
private int boxWidth = 30;
private int boxHeight = 20;

protected void repaint() {
this.getChildren().clear();

int x = startingX + 10;
int y = startingY + 40;

// Display array
x = startingX + 10;
getChildren().add(new Text(x - 15, y + 55, "Array "));
x += 20;
getChildren().add(new Text(x + pivot * boxWidth, y + 120, "Pivot"));
drawArrowLine(x + 15 + pivot * boxWidth, y + 100, x + 15 + pivot * boxWidth, y + 40 + boxHeight);

getChildren().add(new Text(x - 14 + leftP * boxWidth, startingY, "Left Pointer"));
drawArrowLine(x + 15 + leftP * boxWidth, startingY, x + 15 + leftP * boxWidth, y + 40);

getChildren().add(new Text(x - 14 + rightP * boxWidth, startingY, "Right Pointer"));
drawArrowLine(x + 15 + rightP * boxWidth, startingY, x + 15 + rightP * boxWidth, y + 40);

for (int k = 0; k < array.length; k++) {
Rectangle rectangle = new Rectangle(x, y + 40, boxWidth, boxHeight);
rectangle.setFill(Color.WHITE);
rectangle.setStroke(Color.BLACK);
getChildren().add(rectangle);
if (array[k] != 0) {
getChildren().add(new Text(x + 5, y + 55, array[k] + ""));
}
x = x + boxWidth;
}
}

public void drawArrowLine(double x1, double y1, double x2, double y2) {
getChildren().add(new Line(x1, y1, x2, y2));

// find slope of this line
double slope = ((((double) y1) - (double) y2)) / (((double) x1) - (((double) x2)));

double arctan = Math.atan(slope);

// This will flip the arrow 45 off of a
// perpendicular line at pt x2
double set45 = 1.57 / 2;

// arrows should always point towards i, not i+1
if (x1 < x2) {
// add 90 degrees to arrow lines
set45 = -1.57 * 1.5;
}

// set length of arrows
int arrlen = 15;

// draw arrows on line
getChildren().add(new Line(x2, y2, (x2 + (Math.cos(arctan + set45) * arrlen)),
((y2)) + (Math.sin(arctan + set45) * arrlen)));

getChildren().add(new Line(x2, y2, (x2 + (Math.cos(arctan - set45) * arrlen)),
((y2)) + (Math.sin(arctan - set45) * arrlen)));
}
}
}

最佳答案

如果没有协程,你就无法单步执行递归函数(Java 中有几个用于协程的库,但其中大多数都调整了 JVM 或字节码,所以不是真正的标准内容)。

你基本上有两个选择:

  • 反转职责,如果可以的话:让 QuickSort 调用动画重绘,而不是调用单步 QuickSort 的动画。

  • 或者,为了实现执行单步快速排序的函数,您将需要展开递归,这将要求您使用显式的 from- 堆栈来跟踪“深度”对。堆栈以“整个数组”开始;在函数的顶部,您弹出一个状态并对该范围进行排序;当您需要递归时,您将较小的范围插入堆栈;当堆栈为空时,您就完成了。

关于java - 快速排序算法的动画,我们在Stack Overflow上找到一个类似的问题: https://stackoverflow.com/questions/36961980/

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