Question

目前我正在使用java构建游戏拼图。首先我的算法是创建按钮阵列2d，在这些按钮上设置图像并将它们添加到面板（网格布局3 * 3）。我的问题是如何将这些按钮上的每一张图像与原始图像进行比较？我在下面的代码中添加按钮到Jpanel。

private void set() throws Exception {
    position = new int[row][col];
    lstBtn = new JButton[row][col];
    count = new int[row * col];
    arr=new ArrayList<>(9);

    panelContainer.setLayout(null);
    panelContainer.setLayout(new GridLayout(row, col));
    BufferedImage img;
    int numCount = 0;
    int posNum=0; 

    orgImg=new BufferedImage[row][col];
    for (int i = 0; i < row; i++) {
         for (int j = 0; j < col; j++) {

           position[i][j] = posNum++;

            if (i ==0 & j == 0) {
           // For blank Button
                lstBtn[i][j] = new JButton();
                lstBtn[i][j].setBackground(Color.WHITE);
               // panelContainer.add(lstBtn[i][j]);
                lstBtn[i][j].setVisible(true);
                arr.add(lstBtn[i][j]);
                count[0] = numCount;
            }
            else {
                   numCount++;
                lstBtn[i][j] = new JButton();
                lstBtn[i][j].addActionListener(this);
             img = image_cutting.getSubImage(j,i);//160, 116 //120, 87,
             orgImg[j][i]=img;
              lstBtn[i][j].setIcon(new ImageIcon(orgImg[j][i].getScaledInstance(160, 116, BufferedImage.SCALE_SMOOTH)));
                lstBtn[i][j].setHorizontalTextPosition(JButton.CENTER);
                lstBtn[i][j].setVerticalTextPosition(JButton.CENTER);
                lstBtn[i][j].setFont(new Font("Dialog",2,20));
                lstBtn[i][j].setForeground(Color.BLUE);
                 lstBtn[i][j].setText("" + numCount); 
                arr.add(lstBtn[i][j]);
             //  panelContainer.add(lstBtn[i][j]);
               lstBtn[i][j].setVisible(true);
            }
            if (numCount == 0) {
                continue;
            }
            count[numCount] = numCount;
        }

此图像是以匹配图像滑动的拼图样本

Answer 1

这是一个概念性的想法，需要一些充实，但是。从一个单一的想法开始，它持有它的顺序＆＃34;顺序＆＃34;或＆＃34;索引＆＃34;在拼图和它的图像

 ArrayAdapter(Context context, int resource)
Constructor
    ArrayAdapter(Context context, int resource, int textViewResourceId)
Constructor
    ArrayAdapter(Context context, int resource, T[] objects)
Constructor
    ArrayAdapter(Context context, int resource, int textViewResourceId, T[] objects)
Constructor
    ArrayAdapter(Context context, int resource, List<T> objects)
Constructor
    ArrayAdapter(Context context, int resource, int textViewResourceId, List<T> objects)
Constructor

接下来，将其包装在某种模型中。它应该管理碎片，提供对它们的访问，并可能提供一个＆＃34;被命令＆＃34;检查以确定碎片是否有序。

此示例获取源图像，所需的cols /行数，并将图像切片并切成碎片。然后它将这些碎片混洗并添加一块空白块代替第一块（索引为public class PuzzelPiece { private int index; private Image img; public PuzzelPiece(int index, Image img) { this.index = index; this.img = img; } public Image getImage() { return img; } public int getIndex() { return index; } }）

它提供了一个简单的public class Puzzle { private List<PuzzelPiece> pieces; public Puzzle(BufferedImage source, int cols, int rows) { int rowHeight = source.getHeight() / rows; int colWidth = source.getWidth() / cols; pieces = new ArrayList<>(25); int index = 0; for (int row = 0; row < rows; row++) { for (int col = 0; col < cols; col++) { Image img = source.getSubimage(col * colWidth, row * rowHeight, colWidth, rowHeight); pieces.add(new PuzzelPiece(index++, img)); } } pieces.remove(0); Collections.shuffle(pieces); pieces.add(new PuzzelPiece(0, null)); // Blank } public boolean isOrdered() { // Instead of sorting the list like this each time // you could just maintain two lists to start with // one ordered and one shuffled List<PuzzelPiece> ordered = new ArrayList<>(pieces); Collections.sort(ordered, new Comparator<PuzzelPiece>() { @Override public int compare(PuzzelPiece o1, PuzzelPiece o2) { return o1.getIndex() - o2.getIndex(); } }); boolean isOrdered = true; for (int index = 0; index < ordered.size(); index++) { if (ordered.get(index) != pieces.get(index)) { isOrdered = false; break; } } return isOrdered; } public int size() { return pieces.size(); } public PuzzelPiece getPieceAt(int index) { return pieces.get(index); } public void swap(PuzzelPiece piece, PuzzelPiece with) { int pieceIndex = pieces.indexOf(piece); int withIndex = pieces.indexOf(with); pieces.set(pieceIndex, with); pieces.set(withIndex, piece); } }方法来交换片段和简单的swap方法来检查列表是否有序

这是未经测试的，并且按照＆＃34;并且意在推动这个想法，提供一个实现

如何建立游戏拼图3 * 3？

1 个答案: