递归函数

import sys
# sys.setrecursionlimit(1200)  # 修改递归层数限制

count = 0
def func():
     global count
     count += 1
     print(count)
     func()
     print(456)

func()


# RecursionError：递归错误
# 官网规定的递归的最大深度1000层：为了节省内存空间，不让用户无限使用内存空间
# 1.递归要尽量控制次数，如果需要很多层递归才能解决问题，不适合递归解决
# 2.循环和递归的关系：
        # 递归不是万能的
        # 递归比起循环来说更占用内存


# 你的递归函数 必须要停下来
count = 0
def func():
     global count
     count += 1
     print(count)
     if count == 3:
         return
     func()
     print(456)


func()


# 函数的调用
# 函数的参数
# 函数的返回值

# 一个递归函数要想结束，必须在函数内写一个return，并且return的条件必须是可达到的


def func(count):
     count += 1
     # print(count)
     if count == 5:
         return 5
     ret = func(count)
     return ret

r = func(1)
print(r)

递归函数相关习题：

# # 1.计算阶乘  5! = 5*4*3*2*1
# # 循环：
# r = 1
# for i in range(5, 0, -1):
#     r *= i
# print(r)
#
# # 递归：
# def func(n):
#
#     if n == 1:
#         return 1
#     ret = n * func(n - 1)
#     return ret
#
#
# print(func(5))


# 2.os模块；查看一个文件夹下的所有文件，这个文件下面还有文件夹
import os

# dir_path = 'F:\Lnh_Study\lnh_01_基础部分\lnh_14_模块的多种导入方式'
# dir_ls = os.listdir('F:\Lnh_Study\lnh_01_基础部分\lnh_14_模块的多种导入方式')
# print(dir_ls)
#
# count = len(dir_ls)
# def func(path, dir_ls):
#     global count
#     if count == 0:
#         return
#     for name in dir_ls:
#         count -= 1
#         stats = os.path.isdir(path + '/'+ name)
#
#         if stats and name != '__pycache__':
#             new_path = path + '/' + name
#             print(new_path)
#             new_ls = os.listdir(new_path)
#             print(new_ls)
#             func(new_path, new_ls)
#
#
# func(dir_path, dir_ls)


# 3.os模块；计算一个文件夹下所有文件的大小，这个文件下面还有文件夹，不能用walk
# dir_path = 'F:\Lnh_Study\lnh_01_基础部分\lnh_14_模块的多种导入方式'
# dir_ls = os.listdir(dir_path)
# print(dir_ls)
# def func(path, dir_ls):
#     file_size = 0
#     for name in dir_ls:
#         stats = os.path.isdir(os.path.join(path, name))
#         if stats:
#             new_path = os.path.join(path, name)
#             print(new_path)
#             new_ls = os.listdir(new_path)
#             print(new_ls)
#             ret = func(new_path, new_ls)
#             file_size += ret
#         else:
#             file_size += os.path.getsize(os.path.join(path, name))
#
#     return file_size
#
# ret = func(dir_path, dir_ls)
# print(f'文件夹大小：{ret}字节')


# 4.计算斐波那契数列  找第100个数
# 循环：1 1 2 3 5 8 13 21
def func(n):
    a = 1
    b = 1
    for i in range(n):
        yield a
        a, b = b, a+b
obj = func(2)
print(list(obj))


# 递归：
def func(n):
    n -= 1
    if n == 0:
        return 1, 1

    ret = func(n)
    return ret[1], (sum(ret))

ret = func(100)
print(ret[0])



# 5.三级菜单 可能是n级...
# 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': {},
# }
#
# 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)