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我正在用 Java 制作俄罗斯方块，并且正在旋转一块。

首先，我只是在旋转条形部件。

我觉得我现在这样做的方式不仅有问题，而且完全是矫枉过正。但我不知道该怎么做。

首先，我有一个 keylistener 设置int[] rotatedCoords为calcRotation("right")... 如果向左旋转，rotationsCounter+=1;则将改为递减。

    if (keycode == KeyEvent.VK_D) {

        int[] rotatedCoords = calcRotation("right");
        rotationsCounter+=1;
        clearCurrPosition();
        rotate(rotatedCoords);
        System.out.println(rotationsCounter);
        if (rotationsCounter == 4) {
            rotationsCounter = 0;
        }
        System.out.println(rotationsCounter);
    }

calcRotation(String right or left) 获取 Piece 中所有 4 个 Tiles 的当前坐标并将它们发送到int[] getRotation(String shape, String direction, int[] current X coords, int[] current Y Coords)

public int[] calcRotation(String direction) {   

        for (int i = 0; i < tile.length; i++) {
            currPositionX[i] = tile[i].getX();
            currPositionY[i] = tile[i].getY();  
            System.out.println(currPositionX[i] + ", " + currPositionY[i]);
        } 
        return getRotation("Bar", direction, currPositionX, currPositionY);
    }

然后 getRotation[] 根据选择的旋转方向（右或左）、形状和旋转计数器（0 度、90 度、180 或 270...）设置新坐标

if (direction == "right") {
      if (shape == "Bar") {
           if (rotationsCounter == 0) {
                    currXs[0] += 1;
                    currYs[0] += -1;

                    currXs[1] += 0; 
                    currYs[1] += 0; 

                    currXs[2] += -1; 
                    currYs[2] += 1; 

                    currXs[3] += -2;    
                    currYs[3] += 2;             

                    rightRotate1 = new int[] {currXs[0], currYs[0], currXs[1], currYs[1], currXs[2], currYs[2], currXs[3], currYs[3]};          
                }       
           if (rotationsCounter == 1) {
               ... etc

然后坐标 ( pieceRotations) 将被适当地设置：

        //handle on left rotations
    if (direction == "right") {
        if (shape == "Bar") {
            if (rotationsCounter == 0) {    
                pieceRotations = rightRotate1;                      
            }       
            if (rotationsCounter == 1) {
                pieceRotations = rightRotate2;      
            }                   
            if (rotationsCounter == 2) {
                pieceRotations = rightRotate3;  
            }   
            if (rotationsCounter == 3) {
                pieceRotations = rightRotate0;  
            }               
        }
    }
    if (direction == "left") {
        if (shape == "Bar") {
            if (rotationsCounter == 0) {    
                pieceRotations = rightRotate3;                      
            }       
            if (rotationsCounter == 1) {
                pieceRotations = rightRotate0;      
            }                   
            if (rotationsCounter == 2) {
                pieceRotations = rightRotate1;  
            }   
            if (rotationsCounter == 3) {
                pieceRotations = rightRotate2;  
            }
        }           
    }
    return pieceRotations;
}

最后，rotate(rotatedCoords) 将调用正确的坐标来旋转所有的瓷砖......

public void rotate(int[] rotatedCoordinates) {
    int counterX = 0, counterY = 1;
    if (movePieceValid()) {
        for (int i = 0; i < tile.length; i++) {
            tile[i].setLocation(rotatedCoordinates[counterX], rotatedCoordinates[counterY]);                
            counterX+=2;
            counterY+=2;
        }   
    } else {
        for (int i = 0; i < tile.length; i++) {
            tile[i].setLocation(currPositionX[i], currPositionY[i]);                
        }   
    }           
}

所以，我目前根据每个形状的当前位置计算新坐标的方法left显然right是矫枉过正。有没有我可以遵循的一般指南来大大简化它？我想不出另一种方法来获得每个形状的位置？

4

1 回答 1

4

只有这么多碎片。试试这个：

public abstract class Piece {
    public abstract void rotate(Direction dir);
}

public class BarPiece extends Piece {
    public void rotate(Direction dir) {
        // implement
    }
}

public class TPiece extends Piece {
   // ...
}

public class LeftSPiece extends Piece {
   // ...
}

看起来有点脏，但一直做数学会很慢，而且因为只有这么多可能的部分......

于 2013-04-08T16:17:18.193 回答