怎样抽像Factory Pattern ？

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怎样抽像Factory Pattern ？5

编俄罗斯方块的时候遇到的问题。
方块有很多种如: Left Dog, Right Dog, Stick, Right L, Left L等等,并且每个方块都有一个旋转方法。
我定义了一个接口Piece，所有的方块都继承这个接口:

public interface Piece {
    public Piece rotate();//Returns a piece representing 
                          //the next rotation of this piece.
}

其中rotate()方法会返回一个旋转的方块。
但是问题来了:

//Represents the piece in the shape of left dog
public class LeftDog implements Piece {
    private index = 0;
    private static LeftDog leftDogs[2];
    static {
        //initializes leftDog[0]
        ......
        //initializes leftDog[1]    
        ......
    }

    public Piece rotate() {
        index++; // This line of code also has to be copied to other type of 
                 // pieces. Which will lead to a "boilerplate code".
        return leftDogs[index]; 
    }
}

我想把上面那行

index++;

给抽象出去，有什么办法吗？
问题补充：
实在是不好意思啊，代码打错啦。实在太对不起各位了。。。。。。。。。。。那个LeftDog类应该是这样的：

//Represents the piece in the shape of left dog      
public class LeftDog implements Piece {      
     
    private int index;      
    private static LeftDog leftDogs[] = new LeftDog[2];      
     
    static {      
        //initializes leftDogs[0]   
        LeftDog leftDog_0 = new LeftDog(0);   
        leftDogs[0] = leftDog_0;      
        ......      
  
        //initializes leftDogs[1]     
        LeftDog leftDog_1 = new LeftDog(1);   
        leftDogs[1] = leftDog_1;      
        ......         
        ......      
    }      
     
    private LeftDog(int index) {   
        this.index = index;   
    };      
     
    public Piece rotate() {      
        int nextRotation = index + 1; // This line of code also has to be copied to other type of       
                 // pieces. Which will lead to a "boilerplate code".      
        return leftDogs[nextRotation % leftDogs.length];       
    }      
     
}

我觉得一个LeftDog只有两种旋转出来的形状:

XX
--XX

和

--X
XX
X

所以我想着把LeftDog做成一个工厂，一调用rotate()就返回其中一个形状。Board类里用到Piece的时候只关心当前的Piece是什么形状的，然后在Board.placePiece()被调用的时候会根据当前的Piece的形状来改变Board的状态。其实LeftDog里的index就是记录当前的LeftDog是横着还是竖着的了。

2008年6月27日 16:13

JohnE
25
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1个答案按时间排序按投票排序

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public interface Piece {
	
	public int getIndex();
	
	public Piece rotate();
}

public abstract class AbstractPiece implements Piece{

	public int index = 0;

}

public class LeftDog extends AbstractPiece{
	
	private static LeftDog leftDogs[]= new LeftDog[2];  
	
	public Piece rotate() {
		return leftDogs[getIndex()];   
	}

	public int getIndex() {
		if(index==2){
			index = 0 ;
			return index;
		}
		else{
			return index;
		}
	}
}

不知道是不是你想要的？

2008年6月27日 19:00