day17 python递归案例(二分查找,三级菜单)
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IT文章
•
2022-08-30 23:53:58
递归函数与三级菜单
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menu = {
'北京': {
'海淀': {
'五道口': {
'soho': {},
'网易': {},
'google': {}
},
'中关村': {
'爱奇艺': {},
'汽车之家': {},
'youku': {},
},
'上地': {
'百度': {},
},
},
'昌平': {
'沙河': {
'老男孩': {},
'北航': {},
},
'天通苑': {},
'回龙观': {},
},
'朝阳': {},
'东城': {},
},
'上海': {
'闵行': {
"人民广场": {
'炸鸡店': {}
}
},
'闸北': {
'火车战': {
'携程': {}
}
},
'浦东': {},
},
'山东': {},
}
menu
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def caidan(dic):
if dic:
for k in dic:
print(k)
city = input('>>>')
if city in dic:
return caidan(dic[city])
else:
return '没找到啊'
else:
return "传入的字典是空的"
print(caidan(menu))
答案一
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def threeLM(dic):
while True:
for k in dic: print(k)
key = input('input>>').strip()
if key == 'b' or key == 'q': return key
elif key in dic.keys() and dic[key]:
ret = threeLM(dic[key])
if ret == 'q': return 'q'
threeLM(menu)
答案二
二分查找算法
二分查只能用在排序过得序列中
l = [2,3,5,10,15,16,18,22,26,30,32,35,41,42,43,55,56,66,67,69,72,76,82,83,88]
def search(num,l,start=None,end=None):
start = start if start else 0
end = end if end is None else len(l) - 1
mid = (end - start)//2 + start
if start > end:
return None
elif l[mid] > num :
return search(num,l,start,mid-1)
elif l[mid] < num:
return search(num,l,mid+1,end)
elif l[mid] == num:
return mid
