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,
唯一比較鬱悶的是老想着用java的語法去學python代碼,這點還需要後面慢慢掌握吧,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 os
import sys
#棋盤模塊
def model(dictionary,serial=False):
if serial:
print('-(初版)井字棋遊戲,輸入棋號進行對戰,')
print('對應棋號爲第一行:a1-a2-a3',end=',')
print('對應棋號爲第二行:b1-b2-b3',end=',')
print('對應棋號爲第三行: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('-遊戲結束-')
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('遊戲結束,' + '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('遊戲結束,' + '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('遊戲結束,' + '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('遊戲結束,' + '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('遊戲結束,' + '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('遊戲結束,' + '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('遊戲結束,' + '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('遊戲結束,' + '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 0
if __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('傳值爲空!')
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('遊戲結束,' + '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('遊戲結束,' + '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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