import sys
# sys.setrecursionlimit(1200)  # 修改遞迴層數限制

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

func()


# RecursionError：遞迴錯誤
# 官網規定的遞迴的最大深度1000層：為了節省記憶體空間，不讓使用者無限使用記憶體空間
# 1.遞迴要儘量控制次數，如果需要很多層遞迴才能解決問題，不適合遞迴解決
# 2.迴圈和遞迴的關係：
        # 遞迴不是萬能的
        #
 
 遞迴比起迴圈來說更佔用記憶體


# 你的遞迴函式 必須要停下來
count = 0
def func():
     global count
     count += 1
     print(count)
     if count == 3:
         return
     func()
     print(456)


func()


# 函式的呼叫
# 函式的引數
# 函式的返回值

# 一個遞迴函式要想結束，必須在函式內寫一個return，並且return的條件必須是可達到的


def func(count):
     count += 1
     # print(count)
 

     if count == 5:
         return 5
     ret = func(count)
     return ret

r = func(1)
print(r)

遞迴函式相關習題：

# # 1.計算階乘  5! = 5*4*3*2*1
# # 迴圈：
# r = 1
# for i in range(5, 0, -1):
#     r *= i
# print(r)
#
# # 遞迴：
# def func(n):
#
#     if n == 1:
#         return 1
#     ret = n * func(n - 1)
#     return ret
 

#
#
# print(func(5))


# 2.os模組；檢視一個資料夾下的所有檔案，這個檔案下面還有資料夾
import os

# dir_path = 'F:\Lnh_Study\lnh_01_基礎部分\lnh_14_模組的多種匯入方式'
# dir_ls = os.listdir('F:\Lnh_Study\lnh_01_基礎部分\lnh_14_模組的多種匯入方式')
# 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模組；計算一個資料夾下所有檔案的大小，這個檔案下面還有資料夾，不能用walk
# dir_path = 'F:\Lnh_Study\lnh_01_基礎部分\lnh_14_模組的多種匯入方式'
# 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'資料夾大小：{ret}位元組')


# 4.計算斐波那契數列  找第100個數
# 迴圈：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))


# 遞迴：
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])



# 5.三級選單 可能是n級...
# menu = {
#     '北京': {
#         '海淀': {
#             '五道口': {
#                 'soho': {},
#                 '網易': {},
#                 'google': {}
#             },
#             '中關村': {
#                 '愛奇藝': {},
#                 '汽車之家': {},
#                 'youku': {},
#             },
#             '上地': {
#                 '百度': {},
#             },
#         },
#         '昌平': {
#             '沙河': {
#                 '老男孩': {},
#                 '北航': {},
#             },
#             '天通苑': {},
#             '回龍觀': {},
#         },
#         '朝陽': {},
#         '東城': {},
#     },
#     '上海': {
#         '閔行': {
#             "人民廣場": {
#                 '炸雞店': {}
#             }
#         },
#         '閘北': {
#             '火車戰': {
#                 '攜程': {}
#             }
#         },
#         '浦東': {},
#     },
#     '山東': {},
# }
#
# 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)