俄羅斯方塊 - 修復硬降碰撞

Question

我正在寫一個方法來處理俄羅斯方塊的方向，並且對編寫硬降落的邏輯感到困惑。只要按下某個鍵就會調用移動方法。

public void move(Direction direction) { 
    if (canMove(direction)) { 
     switch (direction) { 
     case DOWN: 
      row = row + 1; 
      break; 
     case LEFT: 
      col = col - 1; 
      break; 
     case RIGHT: 
      col = col + 1; 
      break; 
     case DROP: 
      for(int i = row; i < Grid.HEIGHT; i ++){ 
       if(!grid.isSet(i,col)){ 
        row = row + 1; 
       }  
      } 
      break; 
     } 
    } 
} 

我的想法是找到最遠的開放空間，該塊可以硬降，它會重複下去行，直到它擊中的是空間很難下降。

編輯：這是我的canMove方法，我已經改變了我的情況下降，即時下降確實奏效，但是，使用下拉鍵時似乎存在碰撞問題。我不明白爲什麼

public boolean canMove(Direction direction) { 
    if (!ableToMove) 
     return false; 

    boolean move = true; 
    // if the given direction is blocked, we can't move 
    // remember to check the edges of the grid 
    switch (direction) { 
    case DOWN: 
     if (row == (Grid.HEIGHT - 1) || grid.isSet(row + 1, col)) 
      move = false; 
     break; 

    case DROP: 
     if (row == (Grid.HEIGHT - 1) || grid.isSet(row + 1, col)) 
      move = false; 
     break; 
    // currently doesn't support checking LEFT or RIGHT 
    // MODIFY so that it correctly returns if it can move left or right 
    case LEFT: 
     if (col == (0) || grid.isSet(row, col-1)) 
      move = false; 
     break; 

    case RIGHT: 
     if (row == (Grid.WIDTH - 1) || grid.isSet(row, col+1)) 
      move = false; 
     break; 


    } 
    return move; 
} 

Answer 1

在DROP情況下要求的變化

case DROP: 
     while(canMove(DOWN)){ 
      row = row + 1; 
     } 
     break;