反转python中的链表

Question

我被要求反转一个以head为参数的参数，其中head是一个链表，例如：1-> 2->
3这是从已经定义的函数返回的，我试图以这种方式实现函数reverse_linked_list：

def reverse_linked_list(head):
    temp = head
    head = None
    temp1 = temp.next
    temp2 = temp1.next
    temp1.next = None
    temp2.next = temp1
    temp1.next = temp
    return temp2

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

    def to_linked_list(plist):
    head = None
    prev = None
    for element in plist:
        node = Node(element)
        if not head:
            head = node
        else:
            prev.next = node
        prev = node
    return head

    def from_linked_list(head):
    result = []
    counter = 0
    while head and counter < 100: # tests don't use more than 100 nodes, so bail if you loop 100 times.
        result.append(head.value)
        head = head.next
        counter += 1
    return result

    def check_reversal(input):
        head = to_linked_list(input)
        result = reverse_linked_list(head)
        assert list(reversed(input)) == from_linked_list(result)

称为：check_reversal([1,2,3])。我编写的用于反转列表的函数具有给定的功能，[3,2,1,2,1,2,1,2,1]并且仅适用于长度为3的列表。如何将其概括为长度为列表的n？

Answer 1

U可以使用mod函数获取每次迭代的余数，并且显然可以帮助反转列表。我想你是R和D团的学生

head=None   
prev=None
for i in range(len):
    node=Node(number%10)
    if not head:
        head=node
    else:
        prev.next=node
    prev=node
    number=number/10
return head