344 lines
9.3 KiB
Markdown
344 lines
9.3 KiB
Markdown
|
# 原创
|
|||
|
|
： Python数据结构（四）——递归
|
|||
|
|
|
|||
|
|
# Python数据结构（四）——递归
|
|||
|
|
|
|||
|
|
# 递归
|
|||
|
|
|
|||
|
|
递归是一种解决问题的方法，将问题分解为更小的子问题，直到得到一个足够小的问题可以被很简单的解决。通常递归涉及函数调用自身。递归允许我们编写优雅的解决方案，解决可能很难编程的问题。
|
|||
|
|
|
|||
|
|
## 计算整数列表和
|
|||
|
|
|
|||
|
|
```
|
|||
|
|
# version1
|
|||
|
|
def list_sum(num_list):
|
|||
|
|
num_sum = 0
|
|||
|
|
for i in num_list:
|
|||
|
|
num_sum += i
|
|||
|
|
return num_sum
|
|||
|
|
print list_sum([1,3,5,7,9])
|
|||
|
|
```
|
|||
|
|
|
|||
|
|
```
|
|||
|
|
25
|
|||
|
|
|
|||
|
|
```
|
|||
|
|
|
|||
|
|
```
|
|||
|
|
# version2:不使用循环
|
|||
|
|
def list_num2(num_list):
|
|||
|
|
if len(num_list)==1:
|
|||
|
|
return num_list[0]
|
|||
|
|
else:
|
|||
|
|
return num_list[0]+list_num2(num_list[1:])
|
|||
|
|
print list_num2([1,3,5,7,9])
|
|||
|
|
|
|||
|
|
|
|||
|
|
```
|
|||
|
|
|
|||
|
|
```
|
|||
|
|
25
|
|||
|
|
|
|||
|
|
```
|
|||
|
|
|
|||
|
|
## 递归三定律
|
|||
|
|
|
|||
|
|
## 整数转任意进制字符
|
|||
|
|
|
|||
|
|
```
|
|||
|
|
def to_str(n,base):
|
|||
|
|
conver_string = "0123456789ABCDEF"
|
|||
|
|
if n < base:
|
|||
|
|
return conver_string[n]
|
|||
|
|
else:
|
|||
|
|
return str(to_str(n//base,base))+conver_string[n%base]
|
|||
|
|
print to_str(1453,16)
|
|||
|
|
```
|
|||
|
|
|
|||
|
|
```
|
|||
|
|
5AD
|
|||
|
|
|
|||
|
|
```
|
|||
|
|
|
|||
|
|
## 栈帧：实现递归
|
|||
|
|
|
|||
|
|
```
|
|||
|
|
from pythonds.basic.stack import Stack
|
|||
|
|
rStack = Stack()
|
|||
|
|
|
|||
|
|
def to_str(n,base):
|
|||
|
|
conver_string = "0123456789ABCDEF"
|
|||
|
|
while n>0:
|
|||
|
|
if n<base:
|
|||
|
|
rStack.push(conver_string[n])
|
|||
|
|
else:
|
|||
|
|
rStack.push(conver_string[n%base])
|
|||
|
|
n //= base
|
|||
|
|
res = ""
|
|||
|
|
while not rStack.isEmpty():
|
|||
|
|
res += str(rStack.pop())
|
|||
|
|
return res
|
|||
|
|
|
|||
|
|
print to_str(1453,16)
|
|||
|
|
```
|
|||
|
|
|
|||
|
|
```
|
|||
|
|
5AD
|
|||
|
|
|
|||
|
|
```
|
|||
|
|
|
|||
|
|
## 可视化递归
|
|||
|
|
|
|||
|
|
```
|
|||
|
|
# turtle 是 Python 所有版本的标准库，插图的工具
|
|||
|
|
import turtle
|
|||
|
|
```
|
|||
|
|
|
|||
|
|
```
|
|||
|
|
myTurtle = turtle.Turtle()
|
|||
|
|
myWin = turtle.Screen()
|
|||
|
|
|
|||
|
|
def drawSpiral(myTurtle,linelen):
|
|||
|
|
if linelen > 0:
|
|||
|
|
# 直走
|
|||
|
|
myTurtle.forward(linelen)
|
|||
|
|
# 右拐
|
|||
|
|
myTurtle.right(90)
|
|||
|
|
# 递归
|
|||
|
|
drawSpiral(myTurtle,linelen)
|
|||
|
|
drawSpiral(myTurtle,100)
|
|||
|
|
# 调用函数 myWin.exitonclick()，这是一个方便的缩小窗口的方法，
|
|||
|
|
# 使乌龟进入等待模式，直到你单击窗口，然后程序清理并退出。
|
|||
|
|
myWin.exitonclick()
|
|||
|
|
```
|
|||
|
|
|
|||
|
|
```
|
|||
|
|
# 绘制分型树
|
|||
|
|
import turtle
|
|||
|
|
def tree(branchlen,t):
|
|||
|
|
if branchlen > 5:
|
|||
|
|
t.forward(branchlen)
|
|||
|
|
t.right(20)
|
|||
|
|
tree(branchlen-15,t)
|
|||
|
|
t.left(40)
|
|||
|
|
tree(branchlen-10,t)
|
|||
|
|
t.right(20)
|
|||
|
|
t.backward(branchlen)
|
|||
|
|
def main():
|
|||
|
|
t = turtle.Turtle()
|
|||
|
|
myWin = turtle.Screen()
|
|||
|
|
t.left(90)
|
|||
|
|
t.up()
|
|||
|
|
t.backward(100)
|
|||
|
|
t.down()
|
|||
|
|
t.color("green")
|
|||
|
|
tree(75,t)
|
|||
|
|
myWin.exitonclick()
|
|||
|
|
main()
|
|||
|
|
```
|
|||
|
|
|
|||
|
|
## 谢尔宾斯基三角形
|
|||
|
|
|
|||
|
|
```
|
|||
|
|
import turtle
|
|||
|
|
|
|||
|
|
def drawTriangle(points,color,myTurtle):
|
|||
|
|
myTurtle.fillcolor(color)
|
|||
|
|
myTurtle.up()
|
|||
|
|
myTurtle.goto(points[0][0],points[0][1])
|
|||
|
|
myTurtle.down()
|
|||
|
|
myTurtle.begin_fill()
|
|||
|
|
myTurtle.goto(points[1][0],points[1][1])
|
|||
|
|
myTurtle.goto(points[2][0],points[2][1])
|
|||
|
|
myTurtle.goto(points[0][0],points[0][1])
|
|||
|
|
myTurtle.end_fill()
|
|||
|
|
|
|||
|
|
def getMid(p1,p2):
|
|||
|
|
return ( (p1[0]+p2[0]) / 2, (p1[1] + p2[1]) / 2)
|
|||
|
|
|
|||
|
|
def sierpinski(points,degree,myTurtle):
|
|||
|
|
colormap = ['blue','red','green','white','yellow',
|
|||
|
|
'violet','orange']
|
|||
|
|
drawTriangle(points,colormap[degree],myTurtle)
|
|||
|
|
if degree > 0:
|
|||
|
|
sierpinski([points[0],
|
|||
|
|
getMid(points[0], points[1]),
|
|||
|
|
getMid(points[0], points[2])],
|
|||
|
|
degree-1, myTurtle)
|
|||
|
|
sierpinski([points[1],
|
|||
|
|
getMid(points[0], points[1]),
|
|||
|
|
getMid(points[1], points[2])],
|
|||
|
|
degree-1, myTurtle)
|
|||
|
|
sierpinski([points[2],
|
|||
|
|
getMid(points[2], points[1]),
|
|||
|
|
getMid(points[0], points[2])],
|
|||
|
|
degree-1, myTurtle)
|
|||
|
|
|
|||
|
|
def main():
|
|||
|
|
myTurtle = turtle.Turtle()
|
|||
|
|
myWin = turtle.Screen()
|
|||
|
|
myPoints = [[-100,-50],[0,100],[100,-50]]
|
|||
|
|
sierpinski(myPoints,3,myTurtle)
|
|||
|
|
myWin.exitonclick()
|
|||
|
|
|
|||
|
|
main()
|
|||
|
|
```
|
|||
|
|
|
|||
|
|
## 汉诺塔游戏
|
|||
|
|
|
|||
|
|
有一个印度教寺庙，将谜题交给年轻的牧师。在开始的时候，牧师们被给予三根杆和一堆 64 个金碟，每个盘比它下面一个小一点。他们的任务是将所有 64 个盘子从三个杆中一个转移到另一个。有两个重要的约束，它们一次只能移动一个盘子，并且它们不能在较小的盘子顶部上放置更大的盘子。牧师日夜不停每秒钟移动一块盘子。当他们完成工作时，传说，寺庙会变成灰尘，世界将消失。
|
|||
|
|
|
|||
|
|
虽然传说是有趣的，你不必担心世界不久的将来会消失。移动 64 个盘子的塔所需的步骤数是 2^64 -1 = 18,446,744,073,709,551,615264-1 = 18,446,744,073,709,551,615。以每秒一次的速度，即584,942,417,355584,942,417,355 年！。
|
|||
|
|
|
|||
|
|
这里是如何使用中间杆将塔从起始杆移动到目标杆的步骤：
|
|||
|
|
|
|||
|
|
```
|
|||
|
|
# 一开始盘子都在起始杆上
|
|||
|
|
# Python 提供了我们需要调用的隐含的栈。
|
|||
|
|
def moveTower(height,fromPole, toPole, withPole):
|
|||
|
|
if height >= 1:
|
|||
|
|
moveTower(height-1,fromPole,withPole,toPole)
|
|||
|
|
moveDisk(fromPole,toPole)
|
|||
|
|
moveTower(height-1,withPole,toPole,fromPole)
|
|||
|
|
def moveDisk(fp,tp):
|
|||
|
|
print("moving disk from",fp,"to",tp)
|
|||
|
|
moveTower(4,'fromPole', 'toPole', 'withPole')
|
|||
|
|
```
|
|||
|
|
|
|||
|
|
```
|
|||
|
|
('moving disk from', 'fromPole', 'to', 'withPole')
|
|||
|
|
('moving disk from', 'fromPole', 'to', 'toPole')
|
|||
|
|
('moving disk from', 'withPole', 'to', 'toPole')
|
|||
|
|
('moving disk from', 'fromPole', 'to', 'withPole')
|
|||
|
|
('moving disk from', 'toPole', 'to', 'fromPole')
|
|||
|
|
('moving disk from', 'toPole', 'to', 'withPole')
|
|||
|
|
('moving disk from', 'fromPole', 'to', 'withPole')
|
|||
|
|
('moving disk from', 'fromPole', 'to', 'toPole')
|
|||
|
|
('moving disk from', 'withPole', 'to', 'toPole')
|
|||
|
|
('moving disk from', 'withPole', 'to', 'fromPole')
|
|||
|
|
('moving disk from', 'toPole', 'to', 'fromPole')
|
|||
|
|
('moving disk from', 'withPole', 'to', 'toPole')
|
|||
|
|
('moving disk from', 'fromPole', 'to', 'withPole')
|
|||
|
|
('moving disk from', 'fromPole', 'to', 'toPole')
|
|||
|
|
('moving disk from', 'withPole', 'to', 'toPole')
|
|||
|
|
|
|||
|
|
```
|
|||
|
|
|
|||
|
|
## 动态规划
|
|||
|
|
|
|||
|
|
假设你是一个自动售货机制造商的程序员。你的公司希望通过给每个交易最少硬币来简化工作。假设客户放入 1 美元的钞票并购买 37 美分的商品。你可以用来找零的最小数量的硬币是多少？
|
|||
|
|
|
|||
|
|
假设你的公司决定在埃尔博尼亚部署自动贩卖机，除了通常的 1，5，10 和 25 分硬币，他们还有一个 21 分硬币 。
|
|||
|
|
|
|||
|
|
```
|
|||
|
|
def recMC(coinValueList,change):
|
|||
|
|
minCoins = change
|
|||
|
|
if change in coinValueList:
|
|||
|
|
return 1
|
|||
|
|
else:
|
|||
|
|
for i in [c for c in coinValueList if c <= change]:
|
|||
|
|
numCoins = 1 + recMC(coinValueList,change-i)
|
|||
|
|
if numCoins < minCoins:
|
|||
|
|
minCoins = numCoins
|
|||
|
|
return minCoins
|
|||
|
|
print recMC([1,5,10,25],63)
|
|||
|
|
```
|
|||
|
|
|
|||
|
|
```
|
|||
|
|
6
|
|||
|
|
|
|||
|
|
```
|
|||
|
|
|
|||
|
|
这种算法是非常低效的。事实上，它需要 67,716,925 个递归调用来找到 4 个硬币的最佳解决 63 美分问题的方案。
|
|||
|
|
|
|||
|
|
减少我们工作量的关键是记住一些过去的结果，这样我们可以避免重新计算我们已经知道的结果。一个简单的解决方案是将最小数量的硬币的结果存储在表中。
|
|||
|
|
|
|||
|
|
```
|
|||
|
|
def recMC(coinValueList,change,knownResults):
|
|||
|
|
minCoins = change
|
|||
|
|
if change in coinValueList:
|
|||
|
|
knownResults[change] = 1
|
|||
|
|
return 1
|
|||
|
|
elif knownResults[change] > 0:
|
|||
|
|
return knownResults[change]
|
|||
|
|
else:
|
|||
|
|
for i in [c for c in coinValueList if c <= change]:
|
|||
|
|
numCoins = 1 + recMC(coinValueList,change-i,knownResults)
|
|||
|
|
if numCoins < minCoins:
|
|||
|
|
minCoins = numCoins
|
|||
|
|
knownResults[change] = minCoins
|
|||
|
|
return minCoins
|
|||
|
|
print recMC([1,5,10,25],63,[0]*64)
|
|||
|
|
```
|
|||
|
|
|
|||
|
|
```
|
|||
|
|
6
|
|||
|
|
|
|||
|
|
```
|
|||
|
|
|
|||
|
|
这个修改的算法减少了我们需要为四个硬币递归调用的数量，63美分问题只需 221 次调用！
|
|||
|
|
|
|||
|
|
使用动态规划算法:
|
|||
|
|
|
|||
|
|
```
|
|||
|
|
def dpMakeChange(coinValueList,change,minCoins):
|
|||
|
|
for cents in range(change+1):
|
|||
|
|
coinCount = cents
|
|||
|
|
for j in [c for c in coinValueList if c<= cents]:
|
|||
|
|
if minCoins[cents-j]+1<coinCount:
|
|||
|
|
coinCounts = minCoins[cents-j]+1
|
|||
|
|
minCoins[cents] = coinCount
|
|||
|
|
return minCoins[change]
|
|||
|
|
```
|
|||
|
|
|
|||
|
|
```
|
|||
|
|
# 跟踪使用的硬币
|
|||
|
|
def dpMakeChange(coinValueList,change,minCoins,coinsUsed):
|
|||
|
|
for cents in range(change+1):
|
|||
|
|
coinCount = cents
|
|||
|
|
newCoin = 1
|
|||
|
|
for j in [c for c in coinValueList if c <= cents]:
|
|||
|
|
if minCoins[cents-j] + 1 < coinCount:
|
|||
|
|
coinCount = minCoins[cents-j]+1
|
|||
|
|
newCoin = j
|
|||
|
|
minCoins[cents] = coinCount
|
|||
|
|
coinsUsed[cents] = newCoin
|
|||
|
|
return minCoins[change]
|
|||
|
|
|
|||
|
|
def printCoins(coinsUsed,change):
|
|||
|
|
coin = change
|
|||
|
|
while coin > 0:
|
|||
|
|
thisCoin = coinsUsed[coin]
|
|||
|
|
print thisCoin
|
|||
|
|
coin = coin - thisCoin
|
|||
|
|
|
|||
|
|
def main():
|
|||
|
|
amnt = 63
|
|||
|
|
clist = [1,5,10,21,25]
|
|||
|
|
coinsUsed = [0]*(amnt+1)
|
|||
|
|
coinCount = [0]*(amnt+1)
|
|||
|
|
|
|||
|
|
print("Making change for",amnt,"requires")
|
|||
|
|
print(dpMakeChange(clist,amnt,coinCount,coinsUsed),"coins")
|
|||
|
|
print("They are:")
|
|||
|
|
printCoins(coinsUsed,amnt)
|
|||
|
|
print("The used list is as follows:")
|
|||
|
|
print(coinsUsed)
|
|||
|
|
```
|
|||
|
|
|
|||
|
|
```
|
|||
|
|
main()
|
|||
|
|
```
|
|||
|
|
|
|||
|
|
```
|
|||
|
|
('Making change for', 63, 'requires')
|
|||
|
|
(3, 'coins')
|
|||
|
|
They are:
|
|||
|
|
21
|
|||
|
|
21
|
|||
|
|
21
|
|||
|
|
The used list is as follows:
|
|||
|
|
[1, 1, 1, 1, 1, 5, 1, 1, 1, 1, 10, 1, 1, 1, 1, 5, 1, 1, 1, 1, 10, 21, 1, 1, 1, 25, 1, 1, 1, 1, 5, 10, 1, 1, 1, 10, 1, 1, 1, 1, 5, 10, 21, 1, 1, 10, 21, 1, 1, 1, 25, 1, 10, 1, 1, 5, 10, 1, 1, 1, 10, 1, 10, 21]
|
|||
|
|
|
|||
|
|
```
|