引言：

剛學python好幾天了，從java到python，基礎學起來確實比較容易，語法掌握，基本概念上都比較容易入腦。

唯一比較郁悶的是老想著用java的語法去學python代碼，這點還需要后面慢慢掌握吧，相信學多種語言的你們也有這種經歷吧。

start：開始上代碼了，希望有更好的邏輯思維來寫，自己也是用最笨拙的思路去寫的，如果有可以優化的代碼請各位大神指教

#!/user/bin/python# -*- coding: utf-8 -*-import osimport sys#棋盤模塊def model(dictionary,serial=False): if serial: print(’-(初版)井字棋游戲,輸入棋號進行對戰,’) print(’對應棋號為第一行:a1-a2-a3’,end=’,’) print(’對應棋號為第二行:b1-b2-b3’,end=’,’) print(’對應棋號為第三行: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’] + ’ | ’)#主模塊def main(): dictionary={’a1’:’ ’,’a2’:’ ’,’a3’:’ ’,’b1’:’ ’,’b2’:’ ’,’b3’:’ ’,’c1’:’ ’,’c2’:’ ’,’c3’:’ ’} model(dictionary, True) u1 = ’x’ #用戶1 u2 = ’o’ #用戶2 stepNumber =1 #記錄步數 break_fang = 0 #獲勝者記錄 while(stepNumber<=9): fv = True # 判斷條件2 while fv: num = input(’請用戶u1開始下棋:’) compare=1 #判斷條件1 for x in dictionary: if x.find(num)!=-1:compare=0 if compare ==0: fv=False dictionary[num] = u1 model(dictionary) # 0:繼續 1,用戶1勝,2,用戶2勝 break_fang = forResult(dictionary) if break_fang > 0: break fv =True #清楚狀態 stepNumber+=1 while fv: num1=input(’請用戶u2開始下棋:’) compare = 1 # 判斷條件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(’是否重新開始? yes:no:’) if c.find(’yes’)!=-1: main() else: print(’-游戲結束-’) return#判斷獲勝情況#dictionary:棋盤信息def forResult(dictionary): dicts= dict(dictionary) if dicts[’a1’] == dicts[’a2’] and dicts[’a2’] == dicts[’a3’] and len(dicts[’a3’].strip())>0: print(’游戲結束,’ + ’用戶1-獲勝’ if dicts[’a1’] == ’x’ else ’用戶2-獲勝’) 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(’游戲結束,’ + ’用戶1-獲勝’ if dicts[’a1’] == ’x’ else ’用戶2-獲勝’) 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(’游戲結束,’ + ’用戶1-獲勝’ if dicts[’a1’] == ’x’ else ’用戶2-獲勝’) 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(’游戲結束,’ + ’用戶1-獲勝’ if dicts[’a2’] == ’x’ else ’用戶2-獲勝’) 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(’游戲結束,’ + ’用戶1-獲勝’ if dicts[’a3’] == ’x’ else ’用戶2-獲勝’) 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(’游戲結束,’ + ’用戶1-獲勝’ if dicts[’a3’] == ’x’ else ’用戶2-獲勝’) 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(’游戲結束,’ + ’用戶1-獲勝’ if dicts[’b1’] == ’x’ else ’用戶2-獲勝’) 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(’游戲結束,’ + ’用戶1-獲勝’ if dicts[’c1’] == ’x’ else ’用戶2-獲勝’) return 1 if dicts[’c1’] == ’x’ else 2 else: return 0if __name__ ==’__main__’: main()

補一點更改思路:forResult()的另一種實現,compares()函數:少了6行代碼量。

def compares(dictionary={’’:’’},string=’’): if len(dictionary)>0 | len(string.strip())==0:print(’傳值為空!’) else: axle =(’a1’,’a3’,’b2’,’c1’,’c3’) # 四個角和中間的數特殊判斷 條件1 axle_fang=False #特殊棋號需要多加一種可能性 for x in axle: if string==x:axle_fang=True if axle_fang: #條件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(’游戲結束,’ + ’用戶1-獲勝’ if dictionary[string] == ’x’ else ’用戶2-獲勝’) return 1 if dictionary[string] == ’x’ else 2 # 拆分棋號 splitStr0,splitStr1,普通棋號只需判斷倆種a倆種可能,上下-左右間的位置 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(’游戲結束,’ + ’用戶1-獲勝’ if dictionary[string] == ’x’ else ’用戶2-獲勝’) return 1 if dictionary[string] == ’x’ else 2 else:return 0

end：寫完這些也有九十行代碼量了，總感覺太多了。

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