我已经编写了两个递归函数,它们的功能几乎相同。我试图使递归正确,然后我偶然发现了答案,但是语法使我感到困惑:
def fac(N):
"""
Factorial of N.
"""
################# This makes no goddamn sense to me #################
if N == 1:
return N
####### because N == 1, yet 'return N' is the correct result ########
elif N == 0:
return 1
return N*fac(N-1)
N == 1如何作为退出条件成立,又如何存储fac(N)的结果?函数prod()
与sum()
的模拟功能相同。
def prod(List):
"""
Product of all numbers in a list.
"""
if len(List) == 1:
return List[-1]
return List[-1]*prod(List[:-1])
我不知道最终结果如何存储在List[-1]
中。 python解释器是否以特殊方式理解return arg*func(arg)
?
考虑计算fac(4)
所需的循环:
1: fac(4) -> 4 * fac(3) # It then has to calculate fac(3)
2: fac(3) -> 3 * fac(2) # It then has to calculate fac(2)
3: fac(2) -> 2 * fac(1) # It then has to calculate fac(1)
4: fac(1) -> 1 # Finally we've returned a value - now back up through the loops
3: fac(2) -> 2 * fac(1) == 2 * 1 == 2
2: fac(3) -> 3 * fac(2) == 3 * 2 == 6
1: fac(4) -> 4 * fac(3) == 4 * 6 == 24
第二部分实际上是同一件事-递归向下直到获得一个值,然后将其一直插入到原始请求中。
没什么特别的,但是在这种情况下,请使用打印并浏览。
def fac(N):
""" Factorial of N. """
if N == 1:
return N
elif N == 0:
return 1
return N*fac(N-1)
让我们看看它对fac(3)的工作原理>
# fac(3)
# fac(3) => 3 * fac(3-1)
# fac(3) => 3 * fac(3-1) => 2 * fac(2-1)
# fac(3) => 3 * fac(3-1) => 2 * fac(2-1) => return 1
# fac(3) => 3 * fac(3-1) <= 2 * 1
# fac(3) <= 3 * 2 * 1
# 6