麻烦大虾们帮忙解决下这个，谢谢！！
类似俄罗斯方块的，这里只是一小部分，主要是点击“变形”，无法实现。下面是代码
import javax.swing.*;
import javax.swing.border.Border;
import javax.swing.border.EtchedBorder;
import java.awt.*;
import java.awt.event.*;
public class ErsBlockTest extends JFrame
{
/**
*
*/
private static final long serialVersionUID = 1L;
private JButton
btOne = new JButton("一"),
btTwo = new JButton("T"),
btThree = new JButton("Z"),
btFour = new JButton("L"),
btXuanz=new JButton("变形");
private JPanel jpanel=new JPanel();
private JPanel jpanel2=new JPanel();
private int temp;
JButton button[][]=new JButton[4][4];
JPanel boxPanel=new JPanel(new GridLayout(4, 4));
private void setTopButton()
{
btOne.addActionListener(new ActionListener() {
public void actionPerformed(ActionEvent ae) {
setButtonBackgroud();
int key = 0x8000;
for (int i = 0; i < button.length; i++) {
for (int j = 0; j < button[i].length; j++) {
if((key & STYLES[0][0])!=0)
{
button[i][j].setBackground(Color.red);
}
key >>= 1;
}
}
repaint();
}
});
btTwo.addActionListener(new ActionListener() {
public void actionPerformed(ActionEvent ae) {
setButtonBackgroud();
int key = 0x8000;
for (int i = 0; i < button.length; i++) {
for (int j = 0; j < button[i].length; j++) {
if((key & STYLES[1][0])!=0)
{
button[i][j].setBackground(Color.red);
}
key >>= 1;
}
}
repaint();
}
});
btThree.addActionListener(new ActionListener() {
public void actionPerformed(ActionEvent ae) {
setButtonBackgroud();
int key = 0x8000;
for (int i = 0; i < button.length; i++) {
for (int j = 0; j < button[i].length; j++) {
if((key & STYLES[2][0])!=0)
{
button[i][j].setBackground(Color.red);
}
key >>= 1;
}
}
repaint();
}
});
btFour.addActionListener(new ActionListener() {
public void actionPerformed(ActionEvent ae) {
setButtonBackgroud();
int key = 0x8000;
for (int i = 0; i < button.length; i++) {
for (int j = 0; j < button[i].length; j++) {
if((key & STYLES[3][0])!=0)
{
button[i][j].setBackground(Color.red);
}
key >>= 1;
}
}
repaint();
}
});
jpanel.setLayout(new FlowLayout());
jpanel.add(btOne);
jpanel.add(btTwo);
jpanel.add(btThree);
jpanel.add(btFour);
}
private void setMildButton(){
btXuanz.addActionListener(new ActionListener(){
@Override
public void actionPerformed(ActionEvent ae) {
// TODO Auto-generated method stub
setButtonBackgroud();
int Key=0x8000;
for(int i=0;i<4;i++)
for(int j=0;j<4;j++)
if(temp==STYLES[i][j]){
if(j==3){j=0;temp=STYLES[i][j];break;}
else{j++;temp=STYLES[i][j];break;}
}
for(int n=0;n<button.length;n++){
for(int k=0;k<button[n].length;k++){
if((Key & temp)!=0)
{
button[n][k].setBackground(Color.red);
}
Key>>=1;
}
}
repaint();
}});
jpanel2.add(btXuanz);
}
private void setButtomButton()
{
for(int i=0;i<4;i++)
{
for(int j=0;j<4;j++)
{
button[i][j]=new JButton();
boxPanel.add(button[i][j]);
}
}
}
private void setButtonBackgroud()
{
for(int i=0;i<4;i++)
{
for(int j=0;j<4;j++)
{
button[i][j].setBackground(Color.BLACK);
}
}
}
public static void main(String[] args)
{
ErsBlockTest ebt=new ErsBlockTest();
}
public ErsBlockTest()
{
setLayout(new GridLayout(3, 1));
setTopButton();
setMildButton();
setButtomButton();
add(jpanel);
add(jpanel2);
add(boxPanel);
setSize(300,300);
setVisible(true);
addWindowListener(new WindowAdapter()
{
public void windowClosing(WindowEvent e)
{
System.exit(0);
}
});
}
public final static int[][] STYLES = {// 共28种状态
{0x0f00, 0x4444, 0x0f00, 0x4444}, // 长条型的四种状态
{0x04e0, 0x0464, 0x00e4, 0x04c4}, // 'T'型的四种状态
{0x4620, 0x6c00, 0x4620, 0x6c00}, // 反'Z'型的四种状态
{0x2640, 0xc600, 0x2640, 0xc600}, // 'Z'型的四种状态
{0x6220, 0x1700, 0x2230, 0x0740}, // '7'型的四种状态
{0x6440, 0x0e20, 0x44c0, 0x8e00}, // 反'7'型的四种状态
{0x0660, 0x0660, 0x0660, 0x0660}, // 方块的四种状态
};
}
[解决办法]
代码太多了，自己静下心来慢慢找，变形这个是将方块的各种形状算法封装到一个类中，通过监听不同的按钮去调用不同的算法实现方块的变形。实现方法有多种，你这种算法还挺难的。
