# 計(jì)算n的n次方
def func(n):
 return n**n
print(func(10))
f = lambda n: n**n
print(f(10))

lst = [1,5,3,4,6]
lst2 = sorted(lst)
print(lst) # 原列表不會(huì)改變
print(lst2) # 返回的新列表是經(jīng)過(guò)排序的
dic = {1:'A', 3:'C', 2:'B'}
print(sorted(dic)) # 如果是字典. 則返回排序過(guò)后的key

# 根據(jù)字符串長(zhǎng)度進(jìn)行排序
lst = ["魯班七號(hào)", "程咬金", "安琪拉", "阿珂"]
# 計(jì)算字符串長(zhǎng)度
def func(s):
 return len(s)
print(sorted(lst, key=func))

# 根據(jù)字符串長(zhǎng)度進(jìn)行排序
lst = ["魯班七號(hào)", "程咬金", "安琪拉", "阿珂"]
# 計(jì)算字符串長(zhǎng)度
def func(s):
 return len(s)
print(sorted(lst, key=lambda s: len(s)))
lst = [{"id":1, "name":'魯班', "age":28},
  {"id":2, "name":'安琪拉', "age":16},
  {"id":3, "name":'阿珂', "age":25}]
# 按照年齡對(duì)信息進(jìn)行排序
print(sorted(lst, key=lambda e: e['age']))

lst = [1,2,3,4,5,6,7]
ll = filter(lambda x: x%2==0, lst) # 篩選所有的偶數(shù)
print(ll)
print(list(ll))
lst = [{"id":1, "name":'魯班', "age":18},
  {"id":2, "name":'安琪拉', "age":16},
  {"id":3, "name":'阿珂', "age":17}]
fl = filter(lambda e: e['age'] > 16, lst) # 篩選年齡大于16的數(shù)據(jù)
print(list(fl)) 

def func(e):
 return e*e
mp = map(func, [1, 2, 3, 4, 5])
print(mp)
print(list(mp))

# 計(jì)算兩個(gè)列表相同位置的數(shù)據(jù)的和
lst1 = [1, 2, 3, 4, 5]
lst2 = [2, 4, 6, 8, 10]
print(list(map(lambda x, y: x+y, lst1, lst2)))

def func():
 print("我是遞歸")
 func()
func()

def foo(n):
 print(n)
 n += 1
 foo(n)
foo(1)　

import os
def func(filepath,n):
 files = os.listdir(filepath) # 查案當(dāng)前文件的目錄
 for file in files: # 獲取每一個(gè)文件名
  # 獲取文件路徑
  file_p = os.path.join(filepath,file)
  if os.path.isdir(file_p): # 判斷file是否是一個(gè)文件夾
   print("\t"*n,file)
   func(file_p,n+1)
  else:
   print("\t"*n,file)
func("/Volumes/擴(kuò)展盤(pán)/網(wǎng)站css",0)

# 判斷n是否在lst中出現(xiàn). 如果出現(xiàn)請(qǐng)返回n所在的位置
lst = [22, 33, 44, 55, 66, 77, 88, 99, 101, 238, 345, 456, 567, 678, 789]
# 非遞歸算法
# 使用二分法可以提高效率 前提條件有序序列
n = 88
left = 0
right = len(lst) - 1

while left <= right: # 邊界,當(dāng)右邊比左邊還小的時(shí)候退出循環(huán)
 mid = (left + right) // 2 # 這里必須是整除,應(yīng)為索引沒(méi)有小數(shù)
 if lst[mid] > n:
  right = mid - 1
 if lst[mid] < n:
  left = mid + 1
 if lst[mid] == n:
  print("找到這個(gè)數(shù)")
  break
else:
 print("沒(méi)有這個(gè)數(shù)!")

# 遞歸來(lái)完成二分法
lst = [22, 33, 44, 55, 66, 77, 88, 99, 101, 238, 345, 456, 567, 678, 789]
def func(n,left,right):
 if left <= right:
  mid = (left + right) // 2
  if n > lst[mid]:
   left = mid + 1
   return func(n,left,right) # 遞歸,遞歸入口
  elif n < lst[mid]:
   right = mid - 1
   # 深坑,函數(shù)的返回值返回給調(diào)用者
   return func(n,left,right) # 遞歸
  elif lst[mid] == n:
   # print("找到了")
   return mid
 else:
  print("沒(méi)找到")
  return -1 # 避免返回None

# 找66,左邊界0,右邊界len(lst) - 1
ret = func(66,0,len(lst) - 1)
print(ret)

# 遞歸二分法另一種形式,但是無(wú)法實(shí)現(xiàn)位置計(jì)算
lst = [22, 33, 44, 55, 66, 77, 88, 99, 101, 238, 345, 456, 567, 678, 789]
def func(lst,target):
 left = 0
 right = len(lst) - 1
 if left > right:
  print("沒(méi)有這個(gè)數(shù)")
 middle = (left + right)//2
 if target < lst[middle]:
  return func(lst[:middle],target)
 elif target > lst[middle]:
  return func(lst[middle + 1:],target)
 elif target == lst[middle]:
  print("找到這個(gè)數(shù)了")
func(lst,101)

# 時(shí)間復(fù)雜度最低, 空間復(fù)雜度最低
lst1 = [5,6,7,8]
 lst2 = [0,0,0,0,0,1,1,1,1]
 for el in lst1:
 lst2[el] = 1
 lst2[4] == 1 # o(1)