递归，二分法

1.递归

1.必须有一个明确的结束条件

2.每次进入更深一次的递归时，问题规模相比上次都应有所减少

3.递归效率不高，递归层次过多会导致栈溢出。

import time
def age(n):
    print('====>',n)
    time.sleep(1)
    if n == 1:
        return 10
    else:
        return age(n-1)+2
print(age(5))

#n=1,res=10
#n>1,res=age(n-1)+2




====> 5
====> 4
====> 3
====> 2
====> 1
18

2.二分法

data = [1, 3, 6, 7, 9, 12, 14, 16, 17, 18, 20, 21, 22, 23, 30, 32, 33, 35]

def search(num,data):
    print(data)
    if len(data) > 1:   #二分
        mid_index=int(len(data)/2)
        mid_value=data[mid_index]
        if num > mid_value:
            #num在列表的右边
            data=data[mid_index:]
            search(num,data)
        elif num < mid_value:
            #num在列表的左边
            data=data[:mid_index]
            search(num,data)
        else:
            print('find it')
            return
    else:
        if data[0] == num:
            print('find it')
        else:
            print('not exists')

search(17,data)





[1, 3, 6, 7, 9, 12, 14, 16, 17, 18, 20, 21, 22, 23, 30, 32, 33, 35]
[1, 3, 6, 7, 9, 12, 14, 16, 17]
[9, 12, 14, 16, 17]
[14, 16, 17]
[16, 17]
find it