本文实例为大家分享了python实现简单的井字棋的具体代码，供大家参考，具体内容如下
使用python实现井字棋游戏，没有具体算法，只是用随机下棋简单实现：
import random
board = [[\'+\',\'+\',\'+\'],[\'+\',\'+\',\'+\'],[\'+\',\'+\',\'+\']]
def ma(board):
if isempty(board):
a = random.randint(0, 2)
b = random.randint(0, 2)
if board[a][b] != \'X\' and board[a][b] != \'O\':
print(\"机器走：\")
board[a][b] = \'O\'
oput(board)
else:
ma(board)
else:
print(\"平局\")
def oput(board):
print(\" 0 1 2\")
for i in range(3):
print(i, end=\' \')
for j in range(3):
print(board[i][j], end=\" \")
print(\"\")
def winput(i,j):
if board[i][j] == \'X\':
print(\"human win\")
else:
print(\"machine win\")
return 1
def test(board):
for i in range(3):
for j in range(3):
if board[i][j] != \'+\':
if j == 0:
if board[i][j] == board[i][j + 1] == board[i][j + 2]:
return winput(i,j)
if i == 0:
if board[i][j] == board[i + 1][j] == board[i + 2][j]:
return winput(i,j)
if i == 0 and j == 0:
if board[i][j] == board[i + 1][j + 1] == board[i + 2][j + 2]:
return winput(i,j)
if i == 2 and j == 0:
if board[i][j] == board[i - 1][j + 1] == board[i - 2][j + 2]:
return winput(i,j)
def isempty(board):
for i in range(3):
for j in range(3):
if board[i][j] == \'+\':
return True
return False
def main():
print(\"初始棋盘：\")
oput(board)
flag = 0
t = input(\"human first? Y/N human for X, machine for O\\n\")
if t == \'Y\':
while isempty(board):
print(\"人走： \")
a, b = map(int, input(\"输入落子纵横坐标： a,b \\n\").split(\',\'))
if board[a][b] == \'+\':
board[a][b] = \'X\'
oput(board)
flag = test(board)
if flag == 1:
break
else:
print(\"落子位置不对\")
continue
ma(board)
flag = test(board)
if flag == 1:
break
if isempty(board) == 0 and flag == 0:
print(\"平局\")
break
elif t == \'N\':
while isempty(board):
ma(board)
flag = test(board)
if isempty(board) == 0 and flag == 0:
print(\"平局\")
break
if flag == 1:
break
print(\"人走： \")
a, b = map(int, input(\"输入落子纵横坐标： a,b \\n\").split(\',\'))
if board[a][b] == \'+\':
board[a][b] = \'X\'
oput(board)
flag = test(board)
if flag == 1:
break
else:
print(\"落子位置不对\")
continue
if __name__ == \"__main__\":
main()
结果：
以上就是本文的全部内容，希望对大家的学习有所帮助，也希望大家多多支持自学编程网。
