计算链接的lis中的值的数目

Question

我的countInt函数有问题。此外，它被标记为countINT，我在其中放入'-'作为参数。我测试了我的链接列表的创建，它似乎工作正常。所以我觉得我可以安全地排除这个问题。但是，对于NoneType对象没有属性值错误，我不确定哪里出错了。有人能帮我找出错误并指导我改正错误吗？非常感谢您！在

输出：

Traceback (most recent call last):
  File "<stdin>", line 1, in <module>
  File "linked_list.py", line 63, in <module>
    print countInt(r,1)
  File "linked_list.py", line 25, in countInt
    countInt(head.next,n)
  File "linked_list.py", line 25, in countInt
    countInt(head.next,n)
  File "linked_list.py", line 25, in countInt
    countInt(head.next,n)
  File "linked_list.py", line 25, in countInt
    countInt(head.next,n)
  File "linked_list.py", line 25, in countInt
    countInt(head.next,n)
  File "linked_list.py", line 25, in countInt
    countInt(head.next,n)
  File "linked_list.py", line 25, in countInt
    countInt(head.next,n)
  File "linked_list.py", line 25, in countInt
    countInt(head.next,n)
  File "linked_list.py", line 25, in countInt
    countInt(head.next,n)
  File "linked_list.py", line 25, in countInt
    countInt(head.next,n)
  File "linked_list.py", line 25, in countInt
    countInt(head.next,n)
  File "linked_list.py", line 25, in countInt
    countInt(head.next,n)
  File "linked_list.py", line 23, in countInt
    if head.next.value == n:
AttributeError: 'NoneType' object has no attribute 'value'

预期产量：

^{pr2}$

我的代码：

class Node:
    def __init__(self,value):
        self.next = None
        self.value = value

def createLinkedList(root, node):
    if root.next is None:
        root.next = node
    else:
        createLinkedList(root.next, node)

def countInt(head, n, count= 0):        #create a dictionary with keys as the values of the linked list and count up if the value occurs again
    count = 0
    if head.value is None:
        return None
    else:
        if head.next.value == n:
            count += 1
        countInt(head.next, n, count)
        return count


# root
r = Node(1)

# nodes
a = Node(4)
b = Node(1)
c = Node(5)
d = Node('-')
e = Node(4)
f = Node(1)
g = Node(2)
h = Node('-')
i = Node(8)
j = Node(9)
k = Node(8)
l = Node(3)

createLinkedList(r,a)
createLinkedList(r,b)
createLinkedList(r,c)
createLinkedList(r,d)
createLinkedList(r,e)
createLinkedList(r,f)
createLinkedList(r,g)
createLinkedList(r,h)
createLinkedList(r,i)
createLinkedList(r,j)
createLinkedList(r,k)
createLinkedList(r,l)


print countInt(r,1)
print countInt(r,'-')

Answer 1

换行：

if head.value is None:

到

^{pr2}$

该行的目的是知道列表何时应该停止，此时下一个节点将是None。存储在节点中的值与此无关：最后一个节点仍然有一个值（相反，您可能希望将值None存储在列表的前面）。在

---

作为一个单独的问题，您的函数将始终返回0。行countInt(head.next, n, count)实际上并没有对变量count执行任何操作：Python按值传递整数，因此传递的变量不会递增。即使您修复了这个问题，您总是检查head.next这一事实意味着您将跳过列表中的第一个元素。您应该将函数设置为：

def countInt(head, n):
    count = 0
    if head.value == n:
        count = 1
    if head.next is None:
        return count
    return count + countInt(head.next, n)

这实际上让您摆脱了在列表中向下传递count（这是实现递归的一种笨拙方法）。在