递归表达式解释？(蟒蛇)

Question

我一直在做一些编码练习并遇到了这个我很想理解的解决方案.

问题(我重写了一下,所以不容易搜索):

编写一个接受正参数n的函数,并返回在达到单个数字之前必须将n中的数字相乘的次数.例如:

 f(29) => 2  # Because 2*9 = 18, 1*8 = 8, 
                       # and 8 has only one digit.

 f(777) => 4 # Because 7*7*7 = 343, 3*4*3 = 36,
                       # 3*6 = 18, and finally 1*8 = 8.

 f(5) => 0   # Because 5 is already a one-digit number.

某人的解决方案:

from operator import mul
def f(n):
    return 0 if n<=9 else f(reduce(mul, [int(i) for i in str(n)], 1))+1

我不明白的是表达式末尾的"+1"是如何工作的.对不起,我无法更准确地标题,但我不知道这叫什么.

谢谢!

Answer 1

它为计数加上1,并为乘法值调用函数

让我们拿f(777)=> 4,

It will add one and call f - 343
count = 1
It will add one and call f - 36
count = 2
It will add one and call f -18
count = 3
It will add one and call f - 8

所以结果是4

函数调用看起来像

f(7777)
  =1+f(343)
  =1+(1+f(36))
  =1+(1+(1+f(18)))
  =1+(1+(1+(1+f(8))))
  =1+1+1+1+0 = 4