Question

#finds length of a list recursively, don't understand why it exceeds maximum length, can someone please explain?
def lengtho(xs):

    if(xs == None):
        return 0
    return 1 + lengtho(xs[1:])


lengtho([1,2,3,4,1,4,1,4,1,1])

Answer 1

当代码到达列表末尾（基本情况）时，xs将等于[]（空列表），而不是None。由于[][1:]只返回[]，函数将调用自身直到堆栈溢出。在python中编写这个函数的正确方法是：

def lengthof(xs):
   """Recursively calculate the length of xs.
      Pre-condition: xs is a list"""
   if (xs == []):
     return 0
   else:
     return 1+ lengthof(xs[1:])

Answer 2

如果您正在谈论列表，那么递归的基本情况永远不会达到True值，xs == None将永远是False，可能您想要将其更改为xs == []或只是if not xs

为什么这个递归函数在python中超过了最大长度？

2 个答案: