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 nn += 1print(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 recursionrecursion()File "C:/Users/thinkpad/Envs/daily_test/test_project/递归函数.py", line 25, in recursionrecursion()File "C:/Users/thinkpad/Envs/daily_test/test_project/递归函数.py", line 25, in recursionrecursion()[Previous line repeated 992 more times]File "C:/Users/thinkpad/Envs/daily_test/test_project/递归函数.py", line 24, in recursionprint(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 nn += 1print(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 18elif n > 0 and n < 5:return age(n+1) +2print(age(1))# 或
def age(n):if n == 1:return 18else:return age(n-1)+2print(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 endmid_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, aimelse:return "找不到这个值", aimret = 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 1return 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] += 1if n == 1 or n == 2:lst[0] -= 1return 1, 1else:a, b = fib(n - 1)lst[0] -= 1if lst[0] == 0:return a+breturn b, a+bprint(fib(6))# 结果呈现
8

4.2 阶乘

def fac(n):if n == 1:return 1return 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 keyelif 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

• 转自 ：http://www.cnblogs.com/Eva-J/articles/7205734.html