python入门之井字棋小游戏
引言:
刚学python好几天了,从java到python,基础学起来确实比较容易,语法掌握,基本概念上都比较容易入脑。
唯一比较郁闷的是老想着用java的语法去学python代码,这点还需要后面慢慢掌握吧,相信学多种语言的你们也有这种经历吧。
start:开始上代码了,希望有更好的逻辑思维来写,自己也是用最笨拙的思路去写的,如果有可以优化的代码请各位大神指教
#!/user/bin/python
#-*-coding:utf-8-*-
importos
importsys
#棋盘模块
defmodel(dictionary,serial=False):
ifserial:
print('-(初版)井字棋游戏,输入棋号进行对战,')
print('对应棋号为第一行:a1-a2-a3',end=',')
print('对应棋号为第二行:b1-b2-b3',end=',')
print('对应棋号为第三行:c1-c2-c3')
print(dictionary['a1']+'|'+dictionary['a2']+'|'+dictionary['a3']+'|')
print('-+-+-+-')
print(dictionary['b1']+'|'+dictionary['b2']+'|'+dictionary['b3']+'|')
print('-+-+-+-')
print(dictionary['c1']+'|'+dictionary['c2']+'|'+dictionary['c3']+'|')
#主模块
defmain():
dictionary={'a1':'','a2':'','a3':'','b1':'','b2':'','b3':'','c1':'','c2':'','c3':''}
model(dictionary,True)
u1='x'#用户1
u2='o'#用户2
stepNumber=1#记录步数
break_fang=0#获胜者记录
while(stepNumber<=9):
fv=True#判断条件2
whilefv:
num=input('请用户u1开始下棋:')
compare=1#判断条件1
forxindictionary:
ifx.find(num)!=-1:compare=0
ifcompare==0:
fv=False
dictionary[num]=u1
model(dictionary)
#0:继续1,用户1胜,2,用户2胜
break_fang=forResult(dictionary)
ifbreak_fang>0:break
fv=True#清楚状态
stepNumber+=1
whilefv:
num1=input('请用户u2开始下棋:')
compare=1#判断条件1
forxindictionary:
ifx.find(num1)!=-1:compare=0
ifcompare==0:
fv=False
dictionary[num1]=u2
model(dictionary)
break_fang=forResult(dictionary)
ifbreak_fang>0:break
stepNumber+=1
gameover(break_fang)
#退出下棋
defgameover(break_fang):
c=input('是否重新开始?yes:no:')
ifc.find('yes')!=-1:
main()
else:
print('-游戏结束-')
return
#判断获胜情况
#dictionary:棋盘信息
defforResult(dictionary):
dicts=dict(dictionary)
ifdicts['a1']==dicts['a2']anddicts['a2']==dicts['a3']andlen(dicts['a3'].strip())>0:
print('游戏结束,'+'用户1-获胜'ifdicts['a1']=='x'else'用户2-获胜')
return1ifdicts['a1']=='x'else2
elifdicts['a1']==dicts['b2']anddicts['b2']==dicts['c3']andlen(dicts['c3'].strip())>0:
print('游戏结束,'+'用户1-获胜'ifdicts['a1']=='x'else'用户2-获胜')
return1ifdicts['a1']=='x'else2
elifdicts['a1']==dicts['b1']anddicts['b1']==dicts['c1']andlen(dicts['c1'].strip())>0:
print('游戏结束,'+'用户1-获胜'ifdicts['a1']=='x'else'用户2-获胜')
return1ifdicts['a1']=='x'else2
elifdicts['a2']==dicts['b2']anddicts['b2']==dicts['c2']andlen(dicts['c2'].strip())>0:
print('游戏结束,'+'用户1-获胜'ifdicts['a2']=='x'else'用户2-获胜')
return1ifdicts['a2']=='x'else2
elifdicts['a3']==dicts['b3']anddicts['b3']==dicts['c3']andlen(dicts['c3'].strip())>0:
print('游戏结束,'+'用户1-获胜'ifdicts['a3']=='x'else'用户2-获胜')
return1ifdicts['a3']=='x'else2
elifdicts['a3']==dicts['b2']anddicts['b3']==dicts['c1']andlen(dicts['c1'].strip())>0:
print('游戏结束,'+'用户1-获胜'ifdicts['a3']=='x'else'用户2-获胜')
return1ifdicts['a3']=='x'else2
elifdicts['b1']==dicts['b2']anddicts['b2']==dicts['b3']andlen(dicts['b3'].strip())>0:
print('游戏结束,'+'用户1-获胜'ifdicts['b1']=='x'else'用户2-获胜')
return1ifdicts['b1']=='x'else2
elifdicts['c1']==dicts['c2']anddicts['c2']==dicts['c3']andlen(dicts['c3'].strip())>0:
print('游戏结束,'+'用户1-获胜'ifdicts['c1']=='x'else'用户2-获胜')
return1ifdicts['c1']=='x'else2
else:
return0
if__name__=='__main__':
main()
补一点更改思路:forResult()的另一种实现,compares()函数:少了6行代码量。
defcompares(dictionary={'':''},string=''):
iflen(dictionary)>0|len(string.strip())==0:print('传值为空!')
else:
axle=('a1','a3','b2','c1','c3')#四个角和中间的数特殊判断条件1
axle_fang=False#特殊棋号需要多加一种可能性
forxinaxle:
ifstring==x:axle_fang=True
ifaxle_fang:#条件1
ifdictionary['a1']==dictionary['b2']anddictionary['b2']==dictionary['c3']anddictionary['c3'].strip()!=''\
ordictionary['a3']==dictionary['b2']anddictionary['b2']==dictionary['c1']anddictionary['c1'].strip()!='':
print('游戏结束,'+'用户1-获胜'ifdictionary[string]=='x'else'用户2-获胜')
return1ifdictionary[string]=='x'else2
#拆分棋号splitStr0,splitStr1,普通棋号只需判断俩种a俩种可能,上下-左右间的位置
splitStr0,splitStr1=string[0],string[1]
print(splitStr0+":"+splitStr1)
ifdictionary[splitStr0+'1']==dictionary[splitStr0+'2']anddictionary[splitStr0+'2']==dictionary[splitStr0+'3']\
ordictionary['a'+splitStr1]==dictionary['b'+splitStr1]anddictionary['b'+splitStr1]==dictionary['c'+splitStr1]:
print('游戏结束,'+'用户1-获胜'ifdictionary[string]=='x'else'用户2-获胜')
return1ifdictionary[string]=='x'else2
else:return0
end:写完这些也有九十行代码量了,总感觉太多了。
控制台打印:
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