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【简单的井字棋(Tic-Tac-Toe)游戏的Python代码实现】

2024/10/25 0:33:41 来源:https://blog.csdn.net/qq_36253366/article/details/140962919  浏览:    关键词:【简单的井字棋(Tic-Tac-Toe)游戏的Python代码实现】

以下是一个简单的井字棋(Tic-Tac-Toe)游戏的Python代码实现。这个游戏允许两位玩家轮流在3x3的棋盘上放置自己的标记(X和O),首先连成一线(横、竖、斜)的玩家获胜。

# Tic-Tac-Toe Game in Pythondef print_board(board):for row in board:print(" | ".join(row))print("-" * 7)def check_winner(board):# Check rowsfor row in board:if row[0] == row[1] == row[2] != " ":return row[0]# Check columnsfor col in range(3):if board[0][col] == board[1][col] == board[2][col] != " ":return board[0][col]# Check diagonalsif board[0][0] == board[1][1] == board[2][2] != " ":return board[0][0]if board[0][2] == board[1][1] == board[2][0] != " ":return board[0][2]return Nonedef play_game():board = [[" " for _ in range(3)] for _ in range(3)]current_player = "X"game_over = Falsewhile not game_over:print_board(board)print(f"Player {current_player}'s turn.")row = int(input("Enter row (0-2): "))col = int(input("Enter column (0-2): "))if board[row][col] != " ":print("This position is already taken. Try again.")continueboard[row][col] = current_playerwinner = check_winner(board)if winner:print_board(board)print(f"Player {winner} wins!")game_over = Trueelif all(cell != " " for row in board for cell in row):print_board(board)print("It's a tie!")game_over = Trueelse:current_player = "O" if current_player == "X" else "X"if __name__ == "__main__":play_game()

这段代码首先定义了一个print_board函数来打印棋盘的当前状态,然后定义了一个check_winner函数来检查是否有玩家获胜。play_game函数是游戏的主循环,它处理玩家的输入,更新棋盘状态,并检查游戏是否结束。如果游戏结束,它会打印出获胜者或平局的信息。

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