如何在循环下在许多条件下满足循环时使用?＃

Question

我设法让2个玩家转弯，但是第二名球员进入她的答案后停止。我想做时循环，但是我认真地想到了如何做。我想到的是我希望两个玩家继续比赛，直到填补了3个位置或xs的3个位置，但是将其放入代码中，我真的需要帮助。

以下是我的病情，但我希望它继续询问玩家，直到满足这些状况

if list1[0,1,2] == 'x':
print ('Congrats! x won!')
elif (list1 [0] , list2 [0] , list3[0]) == 'x':
print ('Congrats! x won!')
elif (list1 [0] , list2 [1], list3[2]) == 'x':
print ('Congrats! x won!')
elif (list2 [0] , list2 [1], list2[2])== 'x':
print ('Congrats! x won!')
elif (list3 [0] , list3 [0], list3[0]) == 'x':
print ('Congrats! x won!')
elif (list1 [1] , list2 [1], list3[1]) == 'x':
print ('Congrats! x won!')
elif (list1 [2] , list2 [2], list3[2]) == 'x':
print ('Congrats! x won!')
elif (list1 [0] , list2 [1], list3[0]) == 'x':
print ('Congrats! x won!')
elif (list1 [1] , list2 [1], list3[1]) == 'x':
print ('Congrats! x won!')
elif (list1 [2] , list2 [2], list3[2]) == 'x':
print ('Congrats! x won!')
elif (list1 [0] , list2 [0], list3[0]) == 'x':
print ('Congrats! x won!')
elif (list1[0,1,2] == 'o':)
print ('Congrats! o won!')
elif (list1 [0] , list2 [0] ,list3[0]) == 'o':
print ('Congrats! o won!')
elif (list1 [0] , list2 [1], list3[2]) == 'o':
print ('Congrats! o won!')
elif (list2 [0] , list2 [1], list2[2]) == 'o':
print ('Congrats! o won!')
elif (list3 [0] , list3 [0], list3[0]) == 'o':
print ('Congrats! o won!')
elif (list1 [1] , list2 [1], list3[1]) == 'o':
print ('Congrats! o won!')
elif (list1 [2] , list2 [2], list3[2]) == 'o':
print ('Congrats! o won!')
elif (list1 [0] , list2 [1], list3[0]) == 'o':
print ('Congrats! o won!')
elif (list1 [1] , list2 [1], list3[1]) == 'o':
print ('Congrats! o won!')
elif (list1 [2] , list2 [2], list3[2]) == 'o':
print ('Congrats! o won!')
elif (list1 [0] , list2 [0], list3[0]) == 'o':
print ('Congrats! o won!') 
else

Answer 1

将您的董事会作为3个列表的列表（每个有3个元素）更容易。

使用all和any，您可以避免许多不必要的重复：

board = [['x', None, 'o'],
         ['o', 'x', None],
         ['o', None, 'x']
         ]

def three_in_row(board, player):
    return any(all(board[j][i] == player for i in range(3)) for j in range(3))

def three_in_column(board, player):
    return any(all(board[i][j] == player for i in range(3)) for j in range(3))

def three_in_diagonal(board, player):
    return all(board[i][i] == player for i in range(3)) or
        all(board[i][2 - i] == player for i in range(3))

print three_in_row(board, 'x')
# False
print three_in_column(board, 'x')
# False
print three_in_diagonal(board, 'x')
# True

Answer 2

尝试以下：

def check_line(line):
    if line[0]==line[1] and line[1]==line[2]:
        return line[0]
    return ' '
def check_win(board):
    #check horizontals
    for row in board:
        res=check_line(row)
        if res != ' ':
            return res
    #check verticals
    x=0
    for col in board[0]:
        res=check_line([board[y][x] for y in range(0, 3)])
        x+=1
        if res != ' ':
            return res
    #check diagonals
    for diagonal in ( [board[0][0], board[1][1], board[2][2]], [board[0][2], board[1][1], board[2][0]]):
        res = check_line(diagonal)
        if res != ' ':
            return res
    return ' '
def show_winner(list1, list2, list3):
    winner=check_win([list1, list2, list3])
    if winner!=" ":
        print("Congrats! {0} won!".format(winner))
list1=['o', 'x', 'x']
list2=['o', 'x', ' ']
list3=['x', 'o', ' ']
show_winner(list1, list2, list3)