从零开始学Python-Day18-递归函数

Python零基础 木人张 7个月前 (03-22) 88次浏览 0个评论 扫描二维码

函数内部也可以调用其他函数,如果一个函数自己调用了自己,就是递归函数。
我们以阶乘为例1x2x3x4x······x(n-1)xn,用函数f(n)表示,有:
f(n)=n!=1x2x3x4x······x(n-1)xn=(n-1)!xn=f(n-1)xn,有且只有n等于1时,属于特殊情况,需要特殊处理:

>>> def f(n):
	if n==1:
		return 1
	return n*f(n-1)

>>> f(1)
1
>>> f(2)
2
>>> f(10)
3628800

对于f(5),我们可以拆解程序的计算过程如下:

判断n!=0,
f(5)=5*f(4)
f(5)=5*(4*f(3))
f(5)=5*(4*(3*f(2)))
f(5)=5*(4*(3*(2*f(1))))
f(5)=5*(4*(3*(2*1)))
f(5)=5*(4*(3*2))
f(5)=5*(4*6)
f(5)=5*24
f(5)=120

基本所有的递归函数都可以写成循环形式,但从逻辑上递归函数更加清晰、简单。
使用递归函数到防止栈溢出,函数调用是通过栈(stack)这种数据结构实现的,每当进入一个函数调用,栈就会加一层栈帧,每当函数返回,栈就会减一层栈帧。由于栈的大小不是无限的,所以,递归调用的次数过多,会导致栈溢出。我们试试f(100)、f(1000)和f(10000)

>>> f(100)
93326215443944152681699238856266700490715968264381621468592963895217599993229915608941463976156518286253697920827223758251185210916864000000000000000000000000
>>> f(1000)
402387260077093773543702433923003985719374864210714632543799910429938512398629020592044208486969404800479988610197196058631666872994808558901323829669944590997424504087073759918823627727188732519779505950995276120874975462497043601418278094646496291056393887437886487337119181045825783647849977012476632889835955735432513185323958463075557409114262417474349347553428646576611667797396668820291207379143853719588249808126867838374559731746136085379534524221586593201928090878297308431392844403281231558611036976801357304216168747609675871348312025478589320767169132448426236131412508780208000261683151027341827977704784635868170164365024153691398281264810213092761244896359928705114964975419909342221566832572080821333186116811553615836546984046708975602900950537616475847728421889679646244945160765353408198901385442487984959953319101723355556602139450399736280750137837615307127761926849034352625200015888535147331611702103968175921510907788019393178114194545257223865541461062892187960223838971476088506276862967146674697562911234082439208160153780889893964518263243671616762179168909779911903754031274622289988005195444414282012187361745992642956581746628302955570299024324153181617210465832036786906117260158783520751516284225540265170483304226143974286933061690897968482590125458327168226458066526769958652682272807075781391858178889652208164348344825993266043367660176999612831860788386150279465955131156552036093988180612138558600301435694527224206344631797460594682573103790084024432438465657245014402821885252470935190620929023136493273497565513958720559654228749774011413346962715422845862377387538230483865688976461927383814900140767310446640259899490222221765904339901886018566526485061799702356193897017860040811889729918311021171229845901641921068884387121855646124960798722908519296819372388642614839657382291123125024186649353143970137428531926649875337218940694281434118520158014123344828015051399694290153483077644569099073152433278288269864602789864321139083506217095002597389863554277196742822248757586765752344220207573630569498825087968928162753848863396909959826280956121450994871701244516461260379029309120889086942028510640182154399457156805941872748998094254742173582401063677404595741785160829230135358081840096996372524230560855903700624271243416909004153690105933983835777939410970027753472000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000
>>> f(10000)
Traceback (most recent call last):
  File "<pyshell#75>", line 1, in <module>
    f(10000)
  File "<pyshell#68>", line 4, in f
    return n*f(n-1)
  File "<pyshell#68>", line 4, in f
    return n*f(n-1)
  File "<pyshell#68>", line 4, in f
    return n*f(n-1)
  [Previous line repeated 1021 more times]
  File "<pyshell#68>", line 2, in f
    if n==1:
RecursionError: maximum recursion depth exceeded in comparison

如何解决溢出的问题呢?这就需要尾递归来优化。尾递归就是函数返回时调用自身,而且return不能使用表达式。对于编译器或解释器而言,每次递归实际上只使用了一个栈帧,就不会出现溢出了。我们来看下:

>>> def f(n):
	return f2(n, 1)

>>> def f2(n2, m):
	if n2 == 1:
		return m
	return f2(n2 - 1,n2 * m)

>>> f(5)
120

这里相当于建立了一个中间函数,每一次return f2(n2-1 , n2 * m)时,f2里面的参数其实已经计算好了,一直使用的都是

我们输入f(5)
==>f2(5,1)
==>f2(4,5)
==>f2(3,20)
==>f2(2,60)
==>f2(1,120)
==>120

然而,实际上,我们一厢情愿的希望这样可以解决爆栈的问题,但是尾递归依然还是递归函数,不做优化一样会爆栈,因为多数编程语言包括Python都没有针对尾递归做优化,即使我们把函数改为尾递归,该爆栈一样爆栈

>>> f(10000)
Traceback (most recent call last):
  File "<pyshell#106>", line 1, in <module>
    f(10000)
  File "<pyshell#95>", line 2, in f
    return f2(n, 1)
  File "<pyshell#100>", line 4, in f2
    return f2(n2 - 1,n2 * m)
  File "<pyshell#100>", line 4, in f2
    return f2(n2 - 1,n2 * m)
  File "<pyshell#100>", line 4, in f2
    return f2(n2 - 1,n2 * m)
  [Previous line repeated 1020 more times]
  File "<pyshell#100>", line 2, in f2
    if n2 == 1:
RecursionError: maximum recursion depth exceeded in comparison

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