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 -*-import osimport sys#棋盘模块def model(dictionary,serial=False): if serial: 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def main(): 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 while fv: num = input(’请用户u1开始下棋:’) compare=1 #判断条件1 for x in dictionary: if x.find(num)!=-1:compare=0 if compare ==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) if break_fang > 0: break fv =True #清楚状态 stepNumber+=1 while fv: num1=input(’请用户u2开始下棋:’) compare = 1 # 判断条件1 for x in dictionary: if x.find(num1)!=-1:compare=0 if compare == 0: fv=False dictionary[num1] = u2 model(dictionary) break_fang = forResult(dictionary) if break_fang > 0: break stepNumber+=1 gameover(break_fang)#退出下棋def gameover(break_fang): c = input(’是否重新开始? yes:no:’) if c.find(’yes’)!=-1: main() else: print(’-游戏结束-’) return#判断获胜情况#dictionary:棋盘信息def forResult(dictionary): dicts= dict(dictionary) if dicts[’a1’] == dicts[’a2’] and dicts[’a2’] == dicts[’a3’] and len(dicts[’a3’].strip())>0: print(’游戏结束,’ + ’用户1-获胜’ if dicts[’a1’] == ’x’ else ’用户2-获胜’) return 1 if dicts[’a1’]==’x’ else 2 elif dicts[’a1’] == dicts[’b2’] and dicts[’b2’] == dicts[’c3’] and len(dicts[’c3’].strip())>0: print(’游戏结束,’ + ’用户1-获胜’ if dicts[’a1’] == ’x’ else ’用户2-获胜’) return 1 if dicts[’a1’] == ’x’ else 2 elif dicts[’a1’] == dicts[’b1’] and dicts[’b1’] == dicts[’c1’] and len(dicts[’c1’].strip())>0: print(’游戏结束,’ + ’用户1-获胜’ if dicts[’a1’] == ’x’ else ’用户2-获胜’) return 1 if dicts[’a1’] == ’x’ else 2 elif dicts[’a2’] == dicts[’b2’] and dicts[’b2’] == dicts[’c2’] and len(dicts[’c2’].strip())>0: print(’游戏结束,’ + ’用户1-获胜’ if dicts[’a2’] == ’x’ else ’用户2-获胜’) return 1 if dicts[’a2’] == ’x’ else 2 elif dicts[’a3’] == dicts[’b3’] and dicts[’b3’] == dicts[’c3’] and len(dicts[’c3’].strip())>0: print(’游戏结束,’ + ’用户1-获胜’ if dicts[’a3’] == ’x’ else ’用户2-获胜’) return 1 if dicts[’a3’] == ’x’ else 2 elif dicts[’a3’] == dicts[’b2’] and dicts[’b3’] == dicts[’c1’] and len(dicts[’c1’].strip())>0: print(’游戏结束,’ + ’用户1-获胜’ if dicts[’a3’] == ’x’ else ’用户2-获胜’) return 1 if dicts[’a3’] == ’x’ else 2 elif dicts[’b1’] == dicts[’b2’] and dicts[’b2’] == dicts[’b3’] and len(dicts[’b3’].strip())>0: print(’游戏结束,’ + ’用户1-获胜’ if dicts[’b1’] == ’x’ else ’用户2-获胜’) return 1 if dicts[’b1’] == ’x’ else 2 elif dicts[’c1’] == dicts[’c2’] and dicts[’c2’] == dicts[’c3’] and len(dicts[’c3’].strip())>0: print(’游戏结束,’ + ’用户1-获胜’ if dicts[’c1’] == ’x’ else ’用户2-获胜’) return 1 if dicts[’c1’] == ’x’ else 2 else: return 0if __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。

def compares(dictionary={’’:’’},string=’’): if len(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 for x in axle: if string==x:axle_fang=True if axle_fang: #条件1 if dictionary[’a1’]==dictionary[’b2’] and dictionary[’b2’]==dictionary[’c3’] and dictionary[’c3’].strip()!=’’ or dictionary[’a3’]==dictionary[’b2’] and dictionary[’b2’]==dictionary[’c1’]and dictionary[’c1’].strip()!=’’: print(’游戏结束,’ + ’用户1-获胜’ if dictionary[string] == ’x’ else ’用户2-获胜’) return 1 if dictionary[string] == ’x’ else 2 # 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) if dictionary[splitStr0+’1’]==dictionary[splitStr0+’2’] and dictionary[splitStr0+’2’]==dictionary[splitStr0+’3’] or dictionary[’a’+splitStr1]==dictionary[’b’+splitStr1] and dictionary[’b’+splitStr1]==dictionary[’c’+splitStr1]: print(’游戏结束,’ + ’用户1-获胜’ if dictionary[string] == ’x’ else ’用户2-获胜’) return 1 if dictionary[string] == ’x’ else 2 else:return 0

end：写完这些也有九十行代码量了，总感觉太多了。

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