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你能帮忙解决一个小问题吗，我怎样才能让我的 pacman 离开左侧，并在右侧输入相同的 x 位置？我知道我的阵列没有反弹。我的 pacman 在我的阵列中旅行。感谢您的帮助。

class PmanMove extends JPanel {





  private void paintPacman(Graphics2D g) {
    g.setColor( Color.yellow );
    g.fill(new Arc2D.Double(Xpos*cellSize, Ypos*cellSize, pRadius, pRadius, 45, 220, Arc2D.PIE));
    g.setFont(new Font(null, Font.PLAIN, 18));
    g.drawString("SCORE "+score, 50, 50);
    repaint();
  }   

  private void paintBoard(Graphics2D g) {  
    for (int i=0;i<area.length; i++) { 
      for(int j=0; j <area[i].length; j++) {  
        if (area[i][j]==1){
          g.setColor(Color.BLACK);
          g.fillRect(i*cellSize,j*cellSize,cellSize,cellSize);}
        // g.fillRect(i*cellSize,j*cellSize,cellSize,cellSize);}
        if (area[i][j]==2){
          g.setColor(Color.YELLOW);
          g.fillOval(i*cellSize+cellSize/4,j*cellSize+cellSize/4,cellSize-cellSize/2,cellSize-cellSize/2);} 
      }
    }
  }

  public void paintComponent(Graphics g) {
    super.paintComponent(g);
    Graphics2D g2d = (Graphics2D) g;
    paintBoard(g2d);
    paintPacman(g2d);
  }



  public void register() {
    setFocusable(true);
    this.addKeyListener(new KeyAdapter() {
      public void keyPressed(KeyEvent e) {

        if( e.getKeyCode() == e.VK_RIGHT ) {
          if(Xpos<(400/cellSize)) {
            switch (area[Xpos+1][Ypos]) {
              case 4:  Xpos+=speed; break;
              case 3:  Xpos+=speed; break;
              case 2: area[Xpos+1][Ypos]=0; Xpos+=speed; score+=5; break;
              case 0: Xpos+=speed; break;
            }
          }
          if( Xpos+1 > (400/cellSize)) {
            Xpos = 0;
          }
        }

        if( e.getKeyCode() == e.VK_LEFT ) {
          if(Xpos>0) {
            switch (area[Xpos-1][Ypos]) {
              case 4: Xpos-=speed; break;
              case 3: Xpos-=speed; break;
              case 2: area[Xpos-1][Ypos]=0; Xpos-=speed; score+=5; break;
              case 0: Xpos-=speed; break;
            }
          }
          if( Xpos <= 0) {
            Xpos = (getWidth()/(cellSize/2));
          }
        }

        if( e.getKeyCode() == e.VK_UP) {
          if(Ypos>0) {
            switch (area[Xpos][Ypos-1]) {
              case 4: Ypos-=speed; break;
              case 3: Ypos-=speed; break;
              case 2: area[Xpos][Ypos-1]=0; Ypos-=speed; score+=5; break;
              case 0: Ypos-=speed; break;
            }
          }
          if( Ypos <=0 ) {
            Ypos = (getHeight()/(cellSize/2));
          }
        }

        if(e.getKeyCode() == e.VK_DOWN) {
          if(Ypos<400/cellSize) {
            switch (area[Xpos][Ypos+1]) {
              case 4:  Ypos+=speed; break;
              case 3:  Ypos+=speed; break;
              case 2: area[Xpos][Ypos+1]=0; Ypos+=speed; score+=5; break;
              case 0: Ypos+=speed; break;
            }
          }
          if( Ypos >= (getHeight()/(cellSize/2))) {
            Ypos = 0;
          } 
        }
      }
    });                   
    requestFocus();

  }


  public static void main( String args[] ) {

    JFrame fr = new JFrame("PmanMove");
    fr.setDefaultCloseOperation(JFrame.EXIT_ON_CLOSE);
    fr.setBackground(new Color(107,106,100));
    fr.setResizable(false);
    PmanMove pman = new PmanMove();
    fr.setSize(400,400);
    fr.add(pman);
    fr.setVisible(true);  
  }
}
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1 回答 1

2

假设您的数组有N元素并且您在 position i，那么向右移动将是：

(i + 1) % N

这%称为模数。如果 ，您将获得索引 0 i+1 = N，从而避免索引超出界限并落在左侧。

于 2013-11-06T21:10:53.137 回答