题目：原题链接（简单）
解法 | 时间复杂度 | 空间复杂度 | 执行用时 |
Ans 1 (Python) |
|
| 40ms (82.50%) |
Ans 2 (Python) | |||
Ans 3 (Python) |
LeetCode的Python执行用时随缘，只要时间复杂度没有明显差异，执行用时一般都在同一个量级，仅作参考意义。
解法一（情景模拟）：
def tictactoe(self, moves: List[List[int]]) -> str:
A = [0 for _ in range(9)]
B = [0 for _ in range(9)]
step = 0
for move in moves:
if step % 2 == 0:
A[move[0] * 3 + move[1]] = 1
else:
B[move[0] * 3 + move[1]] = 1
step += 1
maybes = {(0, 1, 2), (3, 4, 5), (6, 7, 8), (0, 3, 6), (1, 4, 7), (2, 5, 8), (0, 4, 8), (2, 4, 6)}
for maybe in maybes:
if A[maybe[0]] == 1 and A[maybe[1]] == 1 and A[maybe[2]] == 1:
return "A"
elif B[maybe[0]] == 1 and B[maybe[1]] == 1 and B[maybe[2]] == 1:
return "B"
if step == 9:
return "Draw"
else:
return "Pending"
