# 递归：在调用一个函数的过程中，又间接或者直接调用了该函数，称之为递归调用
# 递推和回溯
# age(5)=age(4)+2
# age(4)=age(3)+2
# age(3)=age(2)+2
# age(2)=age(1)+2
# age(1)=18
# age(n)=age(n-1)+2
# age(1)=18
# def age(n):
# if n == 1:
# return 18
# return age(n-1)+2
#
# res=age(5)
# print(res) # 26
# 总结递归的使用：
# 必须有一个明确的结束条件
# 每次进入更深一层递归时，问题规模相比上次递归都应有所减少
# 递归效率不高
# 二分法
# l=[1,10,30,66,88,99,100,1000,9999,10000] # 从小到大排列的数字列表
# def binary_seach(l,num):
# print(l)
# mid_index=len(l)//2
# if num > l[mid_index]:
# binary_seach(l[mid_index+1:],num)
# elif num < l[mid_index]:
# binary_seach(l[0:mid_index], num)
# else:
# print('find it')
#
# binary_seach(l,99)
'''
[1, 10, 30, 66, 88, 99, 100, 1000, 9999, 10000]
find it
'''
# l=[1,10,30,66,88,99,100,1000,9999,10000] # 从小到大排列的数字列表
# def binary_seach(l,num):
# print(l)
# if len(l) == 0:
# print('not exists')
# return
# mid_index=len(l)//2
# if num > l[mid_index]:
# binary_seach(l[mid_index+1:],num)
# elif num < l[mid_index]:
# binary_seach(l[0:mid_index], num)
# else:
# print('find it')
# binary_seach(l,101)
'''
[1, 10, 30, 66, 88, 99, 100, 1000, 9999, 10000]
[100, 1000, 9999, 10000]
[100, 1000]
[100]
[]
not exists
'''
# binary_seach(l,1000)
'''
[1, 10, 30, 66, 88, 99, 100, 1000, 9999, 10000]
[100, 1000, 9999, 10000]
[100, 1000]
find it
'''
