Question

我正在制作Tic Tac Toe游戏，我确实有有效的代码，到目前为止，它可以满足我的要求。我只是想知道是否有一种方法可以缩短此功能。我的代码如下...

def EnterMove(board):
    move = input("Enter your move (number between 1 - 9): ")

    if move == '1':
        board[0][0] = 'O'
    elif move == '2':
        board[0][1] = 'O'
    elif move == '3':
        board[0][2] = 'O'
    elif move == '4':
        board[1][0] = 'O'
    elif move == '5':
        board[1][1] = 'O'
    elif move == '6':
        board[1][2] = 'O'
    elif move == '7':
        board[2][0] = 'O'
    elif move == '8':
        board[2][1] = 'O'
    elif move == '9':
        board[2][2] = 'O'

#Making the playing board
board = []

for i in range(3):
    row = [Empty for i in range(3)]
    board.append(row)

board[0][0] = '1'
board[0][1] = '2'
board[0][2] = '3'
board[1][0] = '4'
board[1][1] = '5'
board[1][2] = '6'
board[2][0] = '7'
board[2][1] = '8'
board[2][2] = '9'

因此，就像我说的那样，到目前为止一切都可以很好地完成，我只是想知道是否有更简单的方法来构建开发板和构建EnterMove函数。非常感谢。

（注意：播放器移动将是“ O”，而计算机的移动将是“ X”，我将为我们提供与播放器移动功能几乎相同的代码，但只需使用{{1} }来决定计算机的移动方向

Answer 1

您可以使用除法和取模来获取行和列：

def EnterMove(board):
    move = int(input("Enter your move (number between 1 - 9): ")) - 1
    board[move // 3][move % 3] = 'O'

board = []

for i in range(3):
    row = [Empty for i in range(3)]
    board.append(row)

for i in range(9):
    board[i // 3][i % 3] = str(i + 1)

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1 个答案: