python实现四人制扑克牌游戏
本文实例为大家分享了python实现四人制扑克牌游戏的具体代码，供大家参考，具体内容如下
题目：
设计一个简单的四人制扑克牌游戏，能够完成以下功能：
1. 洗牌
2. 发牌
3.自定义规则，在每轮单张出牌时，判定赢家
4.自定义规则，判定最终的赢家
规则简化版：
仅能出单张牌，且出牌时，每个人出的是自己手中牌中刚好能压过上家的最小牌，最先出完的为赢家
import random
from random import choice
flower = ['\u2660','\u2663','\u2665','\u2666']
pai = ['3','4','5','6','7','8','9','10','J','Q','K','A','2']
list = []
list0 = []#储存发的牌
list1 = []
list2 = []
list3 = []
value = []
value0 = []#储存牌代表的值
value1 = []
value2 = []
value3 = []
l00 = []#储存进行升序排序后的牌
l11 = []
l22 = []
l33 = []
for i in flower:
for j in pai:
list.append(i+j)
for i in range(4):
for j in range(13):
value.append(j)
d = dict(zip(list,value))
for i in range(13):
for j in range(4):
if(j == 0):
k = choice(list)#随机选牌
for x in range(len(list)):
if(k == list[x]):#和链表的牌进行匹配，删掉对应项
value0.append(d[k])
list.pop(x)
break
list0.append(k)
if(j == 1):
k = choice(list)
for x in range(len(list)):
if (k == list[x]):
value1.append(d[k])
list.pop(x)
break
list1.append(k)
if (j == 2):
k = choice(list)
for x in range(len(list)):
if (k == list[x]):
value2.append(d[k])
list.pop(x)
break
list2.append(k)
if (j == 3):
k = choice(list)
for x in range(len(list)):
if (k == list[x]):
value3.append(d[k])
list.pop(x)
break
list3.append(k)
d0 = dict(zip(list0,value0))#将每个人的牌转换为字典形式
d1 = dict(zip(list1,value1))
d2 = dict(zip(list2,value2))
d3 = dict(zip(list3,value3))
l0 = sorted(d0.values())#对牌所代表的数字进行排序
l1 = sorted(d1.values())
l2 = sorted(d2.values())
l3 = sorted(d3.values())
#对发给每个人的牌进行排序
for i in range(len(l0)):
for j in list0:
if(l0[i] == d0[j]):
l00.append(j)
break
for i in range(len(l1)):
for j in list1:
if(l1[i] == d1[j]):
l11.append(j)
break
for i in range(len(l2)):
for j in list2:
if(l2[i] == d2[j]):
l22.append(j)
break
for i in range(len(l0)):
for j in list3:
if(l3[i] == d3[j]):
l33.append(j)
break
# y = choice(['0','1','2','3'])
print("第一个人的牌：",l00)
print("第二个人的牌：",l11)
print("第三个人的牌：",l22)
print("第四个人的牌：",l33)
y = random.randint(0,3)
if (y == 0):
y = y + 1
n = l0[0]
l0.pop(0)
elif (y == 1):
y = y + 1
n = l1[0]
l1.pop(0)
elif (y == 2):
y = y + 1
n = l2[0]
l2.pop(0)
elif (y == 3):
y = 0
n = l3[0]
l3.pop(0)
for i in range(13):
if(y == 0):
for j in range(len(l0)):
if(l0[j] > n):
n = l0[j]
l0.pop(j)
if(len(l0) != 0 and n >= l1[len(l1)-1] and n >= l2[len(l2)-1] and n >= l3[len(l3)-1]):#判断是否当前牌中最大牌，若是，则该此人继续出牌
n = l0[0]
l0.pop(0)
break
y = y + 1
if (len(l0) == 0):
print("赢家：第一个人")
break
if(y == 1):
for j in range(len(l1)):
if(l1[j] > n):
n = l1[j]
l1.pop(j)
if (len(l1) != 0 and n >= l0[len(l0) - 1] and n >= l2[len(l2) - 1] and n >= l3[len(l3) - 1]):
n = l1[0]
l1.pop(0)
break
y = y + 1
if (len(l1) == 0):
print("赢家：第二个人")
break
if(y == 2):
for j in range(len(l2)):
if(l2[j] > n):
n = l2[j]
l2.pop(j)
if (len(l2) != 0 and n >= l0[len(l0) - 1] and n >= l1[len(l1) - 1] and n >= l3[len(l3) - 1]):
n = l2[0]
l2.pop(0)
break
y = y + 1
if (len(l2) == 0):
print("赢家：第三个人")
break
if (y == 3):
for j in range(len(l3)):
if (l3[j] > n):
n = l3[j]
l3.pop(j)
if (len(l3) != 0 and n >= l0[len(l0) - 1] and n >= l1[len(l1) - 1] and n >= l2[len(l2) - 1]):
n = l3[0]
l3.pop(0)
break
y = 0
if (len(l3) == 0):
print("赢家：第四个人")
break
#将剩余牌从键值转化成牌
if(len(l0) != 0):
for i in range(len(l0)):
for j in list0:
if(l0[i] == d0[j]):
l0[i] = j
break
if(len(l1) != 0):
for i in range(len(l1)):
for j in list1:
if(l1[i] == d1[j]):
l1[i] = j
break
if(len(l2) != 0):
for i in range(len(l2)):
for j in list2:
if(l2[i] == d2[j]):
l2[i] = j
break
if(len(l3) != 0):
for i in range(len(l3)):
for j in list3:
if(l3[i] == d3[j]):
l3[i] = j
break
print("第一个人的牌：",l0)
print("第二个人的牌：",l1)
print("第三个人的牌：",l2)
print("第四个人的牌：",l3)
您可能感兴趣的文章:python实现扑克牌交互式界面发牌程序python模拟实现分发扑克牌python模拟斗地主发牌python模拟实现斗地主发牌基于Python实现扑克牌面试题
