井字棋的Python实现

井字棋,英文名叫Tic-Tac-Toe,是一种在3*3格子上进行的连珠游戏,和五子棋类似,由于棋盘一般不画边框,格线排成井字故得名。游戏需要的工具仅为纸和笔,然后由分别代表O和X的两个游戏者轮流在格子里留下标记(一般来说先手者为X),任意三个标记形成一条直线,则为获胜。
游戏的难点在于,如何判断连接成了一条线;横、竖、斜三个方向;
代码如下:

# 创建井字棋的程序
def initBoard():
 global board # 调用全局的board
 board = [None] * 3
 print("井字棋:")
 for i in range(len(board)):
  board[i] = ["+ "] * 3


# 打印井字棋的程序
def printBoard():
 global board
 for i in range(len(board)):
  for j in range(len(board[i])):
   print(board[i][j], end=" ")
  print("")


# 开始下棋的程序
def startGame():
 global board
 player = 0
 while isGameContinue():
  if player <= 8:
   if player % 2 == 0:
   # 甲方下棋
   print("== 黑方下棋")
   if not playChess("x"):
    continue
   else:
   # 乙方下棋
   print("== 白方下棋")
   if not playChess("○"):
    continue
   player += 1
  else:
   print("平局")
   break

def playChess(chess):
 # 获取位置
 x = int(input("==  X=")) - 1
 y = int(input("==  Y=")) - 1
 if board[x][y] == "+ ":
  board[x][y] = chess
  printBoard()
  return True # 落子成功
 else:
  print("==  已有棋子 请重新落子\a")
  printBoard()
  return False # 落子失败


def isGameContinue():
 for i in range(len(board)):
  for j in range(len(board[i])):
   if board[i][j] != "+ ":
   # 横向
   if j == 0:
    if board[i][j] == board[i][j + 1] == board[i][j + 2]:
     whoWin(i, j)
     return False
   # 竖向
   if i == 0:
    if board[i][j] == board[i + 1][j] == board[i + 2][j]:
     whoWin(i, j)
     return False
   # 正斜
   if i == 0 and j == 0:
    if board[i][j] == board[i + 1][j + 1] == board[i + 2][j + 2]:
     whoWin(i, j)
     return False
   # 反斜
   if i == 2 and j == 0:
    if board[i][j] == board[i - 1][j + 1] == board[i - 2][j + 2]:
     whoWin(i, j)
     return False
 return True


def whoWin(i, j):
 if board[i][j] == "x":
  print("黑方胜!")
 else:
  print("白方胜!")
 for i in range(3):
  print("win")


class main():
 board = []
 initBoard()
 printBoard()
 startGame()


if __name__ == "__main__":
 main()

你可能感兴趣的:(努力学习人工智能,努力学习java,python,开发语言)