反转链表并像原来一样显示

Question

我成功地实现了一个带有打印列表元素的 Display 函数的单链表。 我创建了一个迭代反向函数，但显示的列表缺少最后一个元素并显示 None 。

我多次回顾我的算法。 有什么我想念的吗？

提前致谢。

class node(object):
    def __init__(self, data=None):
        self.data = data
        self.next = None

class LinkedList(object):
    def __init__(self, head=None):
        self.head = node()

    # append to list
    def append(self, data):
        new_node = node(data)
        current = self.head  # head of the list
        while current.next != None:  # while not last node
            current = current.next  # traverse
        current.next = new_node  # append

    def display(self):
        list = []
        current = self.head
        while current.next != None:
            current = current.next
            list.append(current.data)
        print(list)
        return

    def reverse(self): 
        current = self.head
        prev = None
        while current:
            next_ = current.next
            current.next = prev
            prev = current
            current = next_
        self.head = prev

测试用例：

list = LinkedList()
list.append(0)
list.append(1)
list.append(2)
list.append(3)
list.append(4)
list.display()
list.reverse()
list.display()

输出：

[0, 1, 2, 3, 4]
[3, 2, 1, 0, None]

Answer 1

问题在于，由于节点和链表的构造函数，您的链表以空白开头。

class node(object):
    def __init__(self, data=None):
        self.data = data
        self.next = None
class LinkedList(object):
    def __init__(self, head=None):
        self.head = node()

如果您注意到当您创建一个新的 LinkedList 对象时，您将获得一个没有数据的头部，并且您通过首先获得 self.head.next 来补偿您的打印语句/附加：

current = self.head  # head of the list
    while current.next != None:  # while not last node

所以这意味着当你在最后的反向类中设置 self.head 时，你将头部设置为非空白头部，并在打印中跳过它。

为了弥补这一点，您需要创建一个新的空白头并设置在 prev 旁边：

    def reverse(self):
    current = self.head.next
    prev = None
    while current:
        next_ = current.next
        current.next = prev

        prev = current
        current = next_

    #We create a new blank head and set the next to our valid list
    newHead = node()
    newHead.next = prev
    self.head = newHead

输出是

[0, 1, 2, 3, 4]
[4, 3, 2, 1, 0]