SOS 游戏 tkinter 问题，检查确认 SOS 已在列表范围内进行

Question

我目前正在开发一个程序，该程序允许您选择游戏板的大小，然后允许玩家选择 S 或 O 以实现 SOS 的目标。我已经用 gui 中的按钮列表完成了这项工作。现在我坚持的问题是检查是否已发出 SOS，同时还要保持在我的列表索引内。代码中的打印语句是为了帮助我确定正在调用什么方法，这是我遇到的另一个问题，具体取决于单击了哪个按钮，只有在调用语句时才能确定。

def checksos(i, j):
 
  for i in range (len(board[j])):
    for j in range(len(board[j])):
     
        
        if board[i][j]["text"]=='S':
            
            if not  i >= len(board[j]): 
                print("h")
                if board[j][i-1]["text"] == 'O' and board[j][i-2]["text"]== 'S':
                this will execute if there is a horizontal sos
                  print("found horizontal sos")

        #check if it will go out of boundaries horizontally and vertically
            if (not i >= len(board[j])-2) and (not j >= len(board)-2):
                    print("d")
                    if board[j+1][i+1]["text"] == 'O' and board[j+2][i+2]["text"] == 'S':
                #this will execute if there is a diagonal sos
                     print("found diagonal sos")

        #check if it will go out of boundaries vertically
            if not j >= len(board)-2:
                        print("v")
                        if board[j+1][i]["text"] == 'O' and board[j+2][i]["text"] == 'S':
                #this will execute if there is a vertical sos
                         print("found vertical sos")
            if not j >= len(board)-1:
                        print("v")
                        if board[j+1][i]["text"] == 'O' and board[j+2][i]["text"] == 'S':
                #this will execute if there is a vertical sos
                         print("found vertical sos")
            if not j >= len(board):
                        print("v")
                        if board[j+1][i]["text"] == 'O' and board[j+2][i]["text"] == 'S':
                #this will execute if there is a vertical sos
                         print("found vertical sos")


        elif board[i][j]["text"]=='O':
           
           if not  i >= len(board[j])-2: 
                print("not out of bounds")
                if board[j][i-1]["text"] == 'S' and board[j][i+1]["text"]== 'S':
                #this will execute if there is a horizontal sos
                 print("found horizontal sos")

        #check if it will go out of boundaries horizontally and vertically
                if (not i >= len(board[j])-2) and (not j >= len(board)-2):
                    if board[j+1][i+1]["text"] == 'S' and board[j-1][i-1]["text"] == 'S':
                #this will execute if there is a diagonal sos
                     print("found diagonal sos")

        #check if it will go out of boundaries vertically
                    if not j >= len(board)-2:
                        if board[j+1][i]["text"] == 'S' and board[j-1][i]["text"] == 'S':
                #this will execute if there is a vertical sos
                         print("found vertical sos")

Answer 1

你让这变得比它需要的更难。特别是，我会问为什么你让每个板单元都是一本字典？为什么不将字符本身存储在board[j][i] 中？这样会容易很多。

我在这里所做的是，对于每一行和每一列，将整行或整列转换为单个字符串，然后在字符串中搜索“SOS”。

我也对对角线做同样的事情，但它们有点棘手。如果您考虑我在这里使用的 5x5 网格作为示例，前两个字符串从 (2,0)、(1,0)、(0,0)、( 0,1) 和 (0,2)。后两个字符串收集从 (2,0)、(3,0)、(4,0)、(4,1) 和 (4,2) 开始的 NE 到 SW 对角线。那应该都是对角线。

def checksos(board):
    w = len(board[0])
    h = len(board)

    # Check for horizontals.

    for row in board:
        s = ''.join([cell['text'] for cell in row])
        if 'SOS' in s:
            print("found horizontal sos")

    # Check for verticals.

    for col in range(w):
        s = ''.join([board[i][col]['text'] for i in range(h)])
        if 'SOS' in s:
            print("found vertical sos")

    # Check for diagonals.  There are N-2 diagonals in each direction;
    # the outermost 2 are too short to hold SOS.

    for offset in range(0,w-2):
        # Start from the top and go SE.  If offset is 1, the
        # first string gets 1,0 then 2,1 then 3,2; the other
        # string gets 0,1 then 1,2 then 2,3.
        s1 = []
        s2 = []
        s3 = []
        s4 = []
        for i in range(0,w-offset):
            s1.append( board[i+offset][i]['text'] )
            s2.append( board[i][i+offset]['text'] )
            s3.append( board[i+offset][w-i-1]['text'] )
            s4.append( board[h-i-1][i+offset]['text'] )
        if 'SOS' in ''.join(s1) or 'SOS' in ''.join(s2) or \
           'SOS' in ''.join(s3) or 'SOS' in ''.join(s4):
            print("found diagonal sos")

board = []
for i in range(5):
    board.append( [{'text':' '} for _ in range(5)] )

board[1][1]['text'] = 'S'
board[1][2]['text'] = 'O'
board[1][3]['text'] = 'S'

board[1][4]['text'] = 'S'
board[2][4]['text'] = 'O'
board[3][4]['text'] = 'S'

board[2][2]['text'] = 'S'
board[3][3]['text'] = 'O'
board[4][4]['text'] = 'S'

checksos(board)