目录
- 什么是递归
- 递归例题
- 递归函数与二分查找算法
- 相关练习题
1，什么是递归
- 递归的定义：在一个函数里再调用这个函数本身
- 递归的最大深度默认是：997 - - 是python从内存角度出发做的限制
- 递归的缺点： 占内存
- 递归的优点： 会让代码变简单
1.1 测试递归最大深度
n = 0
def recursion():
global n
n += 1
print(n)
recursion()
recursion()
# 结果呈现
1
2
...
995
996
Traceback (most recent call last):
File "C:/Users/thinkpad/Envs/daily_test/test_project/递归函数.py", line 27, in <module>
recursion()
File "C:/Users/thinkpad/Envs/daily_test/test_project/递归函数.py", line 25, in recursion
recursion()
File "C:/Users/thinkpad/Envs/daily_test/test_project/递归函数.py", line 25, in recursion
recursion()
File "C:/Users/thinkpad/Envs/daily_test/test_project/递归函数.py", line 25, in recursion
recursion()
[Previous line repeated 992 more times]
File "C:/Users/thinkpad/Envs/daily_test/test_project/递归函数.py", line 24, in recursion
print(n)
RecursionError: maximum recursion depth exceeded while calling a Python object
RecursionError: maximum recursion depth exceeded while calling a Python object
递归的错误：超过的递归的最大深度
1.2 修改递归最大深度
import sys
sys.setrecursionlimit(100000)
n = 0
def recursion():
global n
n += 1
print(n)
recursion()
recursion()
# 结果呈现
1
2
...
3924
3925
2，递归例题
2.1 A 比 B 大两岁，B 比 C 大两岁，C 比 D 大两岁，D 比 F 大两岁，F 18岁了，求A多大了？
- 分析规律
age(A) = age(B) + 2
age(B) = age(C) + 2
age(C) = age(D) + 2
age(D) = age(F) + 2
age(F) = 18
- 函数如下
def age(n):
if n == 5:
return 18
elif n > 0 and n < 5:
return age(n+1) +2
print(age(1))
# 或
def age(n):
if n == 1:
return 18
else:
return age(n-1)+2
print(age(5))
# 结果呈现
26
3，递归函数与二分查找算法
3.1 例题
- 如果有这样一个列表，让你从这个列表中找到66的位置，你要怎么做？
l = [2,3,5,10,15,16,18,22,26,30,32,35,41,42,43,55,56,66,67,69,72,76,82,83,88]
print(l.index(66))
# 结果呈现
17
- 不用
index，如何查找
l = [2,3,5,10,15,16,18,22,26,30,32,35,41,42,43,55,56,66,67,69,72,76,82,83,88]
i = 0
for num in l:
if num == 66:
print(i)
i+=1
# 结果呈现
17
3.2 二分查找算法
- 二分查找算法，必须处理有序的列表
3.2.1 有一个列表，从这个列表中找到66的位置
l = [2,3,5,10,15,16,18,22,26,30,32,35,41,42,43,55,56,66,67,69,72,76,82,83,88]
通过 index 方法获取
l = [2,3,5,10,15,16,18,22,26,30,32,35,41,42,43,55,56,66,67,69,72,76,82,83,88]
print(l.index(66))
# 结果呈现
17
通过 for 循环获取
l = [2,3,5,10,15,16,18,22,26,30,32,35,41,42,43,55,56,66,67,69,72,76,82,83,88]
i = 0
for num in l:
if num == 66:
print(i)
i+=1
# 结果呈现
17
通过 二分查找算法 获取
l = [2,3,5,10,15,16,18,22,26,30,32,35,41,42,43,55,56,66,67,69,72,76,82,83,88]
def find(l, aim, start = 0, end=None):
end = len(l) if end is None else end
mid_index = (end - start) // 2 + start # 计算中间值
if start <= end:
if l[mid_index] > aim:
return find(l, aim, start = start, end= mid_index - 1)
elif l[mid_index] < aim:
return find(l, aim, start = mid_index + 1, end=end)
else:
return "找到了",mid_index, aim
else:
return "找不到这个值", aim
ret = find(l, 66)
ret1 = find(l, 44)
print(ret)
print(ret1)
# 结果呈现
('找到了', 17, 66)
('找不到这个值', 44)
4 , 相关练习题
4.1 斐波那契 # 问第N个斐波那契数是多少
# 1, 1, 2, 3, 5, 8
# fib(6) = fib(5) + fib(4)
# fib(5) = fib(4) + fib(3)
# fib(4) = fib(3) + fib(2)
# fib(3) = fib(2) + fib(1)
# fib(2) = 1
# fib(1) = 1
- 使用递归 调用自己两次的方法，效率慢，占内存
def fib(n):
if n == 1 or n ==2:
return 1
return fib(n - 1) + fib(n - 2)
ret = fib(6)
print(ret)
# 结果呈现
8
- 使用递归 调用自己一次的方法
def fib(n, lst = [0]):
lst[0] += 1
if n == 1 or n == 2:
lst[0] -= 1
return 1, 1
else:
a, b = fib(n - 1)
lst[0] -= 1
if lst[0] == 0:
return a+b
return b, a+b
print(fib(6))
# 结果呈现
8
4.2 阶乘
def fac(n):
if n == 1:
return 1
return n * fac(n - 1)
print(fac(6))
# 结果呈现
# 6! 6 * 5 * 4 * 3 * 2 * 1
720
4.3 递归函数与三级菜单
- 三级菜单
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': {},
}
- 递归函数实现三级菜单
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)
- 递归 堆栈实现
l = [menu]
while l:
for key in l[-1]:print(key)
k = input('input>>').strip() # 北京
if k in l[-1].keys() and l[-1][k]:l.append(l[-1][k])
elif k == 'b':l.pop()
elif k == 'q':break