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是否可以使用 boolean 而不是列表以这种方式进行洗牌?

[英]Is it possible to do a shuffle in this manner with a boolean instead of a list?

Here my code (In Java on Eclipse) displays 3 random cards from the a file.在这里,我的代码(在 Eclipse 上的 Java 中)显示来自文件的 3 张随机卡片。 I am trying to get a shuffle button to work and randomly shuffle in 3 new cards.我试图让一个洗牌按钮工作并随机洗牌 3 张新牌。 I used "Collections.shuffle(cards);"我使用了“Collections.shuffle(cards);” and passed it my boolean array but it says I can't because it wants a List<> list.并将我的 boolean 数组传递给它,但它说我不能,因为它想要一个 List<> 列表。 Is it possible to get the shuffle to work with my boolean or would I have to use a List?是否可以让洗牌与我的 boolean 一起使用,还是我必须使用列表?

Here is my code:这是我的代码:

import java.util.Collections;
import java.util.List;

import javafx.application.Application;
import javafx.event.ActionEvent;
import javafx.event.EventHandler;
import javafx.geometry.Pos;
import javafx.scene.Scene;
import javafx.scene.control.Button;
import javafx.scene.image.Image;
import javafx.scene.image.ImageView;
import javafx.scene.layout.BorderPane;
import javafx.scene.layout.GridPane;
import javafx.scene.layout.HBox;
import javafx.stage.Stage;

public class DisplayCards extends Application {
    
    HBox imageViews;

    
    @Override
    public void start(Stage primaryStage) throws Exception {
        GridPane pane = new GridPane();
        pane.setAlignment(Pos.CENTER);
        boolean[] cards = new boolean[52];
        int count = 0;
        while(count <3) {
            int card = (int)(Math.random() * 52);
            
            if(!cards[card]) {
                cards[card] = true;
                pane.add(new ImageView(new Image("card/" + (card) + ".png")), count, 0);
                count++;
            }
        }
        imageViews = new HBox();
        imageViews.setAlignment(Pos.CENTER);
        
        shuffle();
    
        
        Button btnShuffle = new Button("Shuffle");
        btnShuffle.setOnAction(new EventHandler<ActionEvent>() {
            
            public void handle(ActionEvent event) {
                shuffle();
                
            }
        });
        
        BorderPane Bpane = new BorderPane();
        Bpane.setCenter(imageViews);
        Bpane.setBottom(btnShuffle);
        
        Scene scene = new Scene(pane, 250, 150);
        primaryStage.setTitle("Display 4 Cards");
        primaryStage.setScene(scene);
        primaryStage.show();
        
    }

    private void shuffle() {
        Collections.shuffle(cards);
        
    }

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

    

}

You can implement the Fisher–Yates shuffle for an array.您可以为数组实现 Fisher-Yates 洗牌。

private void shuffle(){
    for(int i = cards.length - 1; i > 0; i++){
        int j = java.util.concurrent.ThreadLocalRandom.current().nextInt(i + 1);
        boolean temp = cards[i];
        cards[i] = cards[j];
        cards[j] = temp;
    }
}

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