我正在制作一个简单的井字游戏,用户可以在其中指定网格的大小(列和行的数量)。
我需要创建一个函数,该函数可以检查网格中所有对角线的胜利。
对于网格,我使用的是二维列表,看起来像这样(3x3示例):
grid = [['x', '-', 'o'],
['o', 'x', '-'],
['-', '-', 'x']]
grid[row][col]
这应该是一个成功的情况
我已经为垂直和水平获胜创建了一个检查,但是我不太清楚如何检查对角线。
这是我在代码中检查行赢的方法:
min_win_streak
是最小的连胜纪录(例如,在经典的3x3井字游戏中为3)
def check_rows(grid, min_win_streak):
winner = '-'
for row in grid:
win_streak = 0
for element in row:
if element != '-':
if element == winner:
win_streak += 1
winner = element
if win_streak >= min_win_streak:
return True, winner
else:
win_streak = 1
winner = element
else:
win_streak = 0
winner = '-'
return False, None
我需要check_diags()
函数,返回值应该是:
答案 0 :(得分:0)
您可以使用以下嵌套的for循环检查2x2中的对角线:
for i in range(len(grid)-2):
for j in range(len(grid)-2):
if grid[i][j] == grid[i+1][j+1] == grid[i+2][j+2] and grid[i][j] != '-':
print('Winner')
elif grid[i][j+2] == grid[i+1][j+1] == grid[i+2][j] and grid[i][j+2] != '-':
print('Winner')
根据需要将其插入到具有所需返回值的函数中,以代替打印语句。应该可以轻松地更改为3x3、4x4等。
答案 1 :(得分:0)
感谢您发布代码。我用all
函数替换了win_streak检查循环。下面的代码完成的工作比您想要在庞大的网格上使用的工作还要多,但是它易于理解(易于)和适应,并且足够快地用于go-moku游戏(19x19上连续5个)板)。
我留下了一条跟踪print
语句来说明它正在检查的对角线。您应该能够使其适应NE-SW对角线(check_slash_diag
)。
grid = [['x', '-', 'o', 'o'],
['o', 'x', 'o', '-'],
['-', 'o', 'x', 'x'],
['-', 'x', 'o', '-']]
def check_backslash_diag(grid, min_win_streak):
grid_size = len(grid)
grid_extra = grid_size - min_win_streak # This is the "extra" space in a long line
# This pair of loops will iterate through the upper-left square of side
# grid_extra+1, starting points of any diagonals long enough to contain a win.
for start_row in range(grid_extra+1):
for start_col in range(grid_extra+1):
# Extract a diagonal "row" of length min_win_streak
diag = [grid[start_row+i][start_col+i] for i in range(min_win_streak)]
print(start_row, start_col, diag) # DEBUG: Display checked diagonals
if all(diag[i] == diag[0] for i in range(min_win_streak)):
print(diag[0], "wins on NW-SE diagonal at", start_row, start_col)
check_backslash_diag(grid, 3)
输出:
0 0 ['x', 'x', 'x']
x wins on NW-SE diagonal at 0 0
0 1 ['-', 'o', 'x']
1 0 ['o', 'o', 'o']
o wins on NW-SE diagonal at 1 0
1 1 ['x', 'x', '-']