如何检查井字游戏中的胜利？

Question

我的任务是制作井字游戏。网格由 3 行组成。我开发了核心游戏玩法，这很有效，但由于某种原因，我无法检查是否获胜。我在回合结束时有if any(wins_x) 和if any(wins_o)。

row1 = ['-', '-', '-']
row2 = ['-', '-', '-']
row3 = ['-', '-', '-']
table = [row1, row2, row3]

wins_x = [(row1[0] == 'x' and row2[1] == 'x' and row3[2] == 'x'),
          (row3[0] == 'x' and row2[1] == 'x' and row1[2] == 'x'),
          (row1 == ['x', 'x', 'x']),
          (row2 == ['x', 'x', 'x']),
          (row3 == ['x', 'x', 'x']),
          (row1[0] == 'x' and row2[0] == 'x' and row3[0] == 'x'),
          (row1[1] == 'x' and row2[1] == 'x' and row3[1] == 'x'),
          (row1[2] == 'x' and row2[2] == 'x' and row3[2] == 'x')]

wins_o = [(row1[0] == 'o' and row2[1] == 'o' and row3[2] == 'o'),
          (row3[0] == 'o' and row2[1] == 'o' and row1[2] == 'o'),
          (row1 == ['o', 'o', 'o']),
          (row2 == ['o', 'o', 'o']),
          (row3 == ['o', 'o', 'o']),
          (row1[0] == 'o' and row2[0] == 'o' and row3[0] == 'o'),
          (row1[1] == 'o' and row2[1] == 'o' and row3[1] == 'o'),
          (row1[2] == 'o' and row2[2] == 'o' and row3[2] == 'o')]

Answer 1

这应该可行：

def wincheck(table):
  for x in range(0,len(table)):
    if len([y for y in table[x] if y=="x"])==len(table):
      print(f"win x row {x+1}")
    elif len([y for y in table[x] if y=="o"])==len(table):
      print(f"win o row {x+1}")
  for x in range(0,len(table)):
    if len([y for y in table if y[x]=="x"])==len(table):
      print(f"win x Column {x+1}")
    elif len([y for y in table if y[x]=="o"])==len(table):
      print(f"win o Column {x+1}")
  if len([y for y in range(0,len(table)) if table[y][len(table)-1-y] =="x" ])==len(table) or len([y for y in range(0,len(table)) if table[y][y] =="x" ])==len(table):
    print("diagonal win x")
  elif len([y for y in range(0,len(table)) if table[y][len(table)-1-y] =="o" ])==len(table) or len([y for y in range(0,len(table)) if table[y][y] =="o" ])==len(table):
    print("diagonal win o")
wincheck(table)

无论大小如何，它都应该可以工作，但我绝对知道它适用于 3x3。这个函数基本上是检查每一行、每一列和每一对角线，看看里面是否有"o"或"x"。