Question

在铅笔纸游戏中，2名玩家轮流在3x3方格的棋盘上标记'X'和'O'。成功在垂直，水平或斜条纹上标记3个连续'X'或'O'的玩家赢得了游戏。编写一个确定井字游戏结果的函数。

实施例

>>> tictactoe([('X', ' ', 'O'), 
               (' ', 'O', 'O'), 
               ('X', 'X', 'X') ])
"'X' wins (horizontal)."
>>> tictactoe([('X', 'O', 'X'), 
...            ('O', 'X', 'O'), 
...            ('O', 'X', 'O') ])
'Draw.'
>>> tictactoe([('X', 'O', 'O'), 
...            ('X', 'O', ' '), 
...            ('O', 'X', ' ') ])
"'O' wins (diagonal)."
>>> tictactoe([('X', 'O', 'X'), 
...            ('O', 'O', 'X'), 
...            ('O', 'X', 'X') ])
"'X' wins (vertical)."




def tictactoe(moves):
for r in range(len(moves)):
    for c in range(len(moves[r])):      
        if moves[0][c]==moves[1][c]==moves[2][c]:
            a="'%s' wins (%s)."%((moves[0][c]),'vertical')
        elif moves[r][0]==moves[r][1]==moves[r][2]:
            a="'%s' wins (%s)."%((moves[r][0]),'horizontal')
        elif moves[0][0]==moves[1][1]==moves[2][2]:
            a="'%s' wins (%s)."%((moves[0][0]),'diagonal')
        elif moves[0][2]==moves[1][1]==moves[2][0]:
            a="'%s' wins (%s)."%((moves[0][2]),'diagonal')
        else:
            a='Draw.'
print(a)

我写了这样的代码，我的范围不起作用（我认为）。因为，它将r和c的值取为3，而不是0,1,2,3。那么，请有人帮我这个吗？谢谢

Answer 1

当玩家获胜时，你的循环不会退出。我会尝试这样的事情：

def tictactoe_state(moves):
  for r in range(len(moves)):
    for c in range(len(moves[r])):
      if moves[0][c] == moves[1][c] == moves[2][c]:
        return "'%s' wins (%s)." % (moves[0][c], 'vertical')
      elif moves[r][0] == moves[r][1] == moves[r][2]:
        return "'%s' wins (%s)." % (moves[r][0], 'horizontal')
      elif moves[0][0] == moves[1][1] == moves[2][2]:
        return "'%s' wins (%s)." % (moves[0][0], 'diagonal')
      elif moves[0][2] == moves[1][1] == moves[2][0]:
        return "'%s' wins (%s)." % (moves[0][2], 'diagonal')

  # You still have to make sure the game isn't a draw.
  # To do that, see if there are any blank squares.

  return 'Still playing'

另外，我会移动检查对角线的if语句。他们不依赖于r和c。

Answer 2

试试这个..

def tictactoe(moves):
    for r in range(len(moves)):
        for c in range(len(moves[r])):      
            if moves[0][c]==moves[1][c]==moves[2][c]:
                return "\'%s\' wins (%s)." % ((moves[0][c]),'vertical')
            elif moves[r][0]==moves[r][1]==moves[r][2]:
                return "\'%s\' wins (%s)."%((moves[r][0]),'horizontal')
            elif moves[0][0]==moves[1][1]==moves[2][2]:
                return "\'%s\' wins (%s)."%((moves[0][0]),'diagonal')
            elif moves[0][2]==moves[1][1]==moves[2][0]:
                return "\'%s\' wins (%s)."%((moves[0][2]),'diagonal')
    return 'Draw.'

pyschools，井字游戏

2 个答案: