反轉鏈表並像原來一樣顯示

Question

我成功地實現了一個帶有打印列表元素的 Display 函數的單鏈表。 我創建了一個迭代反向函數，但顯示的列表缺少最后一個元素並顯示 None 。

我多次回顧我的算法。 有什么我想念的嗎？

提前致謝。

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

測試用例：

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

輸出：

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

Answer 1

問題在於，由於節點和鏈表的構造函數，您的鏈表以空白開頭。

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()

如果您注意到當您創建一個新的 LinkedList 對象時，您將獲得一個沒有數據的頭部，並且您通過首先獲得 self.head.next 來補償您的打印語句/附加：

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

所以這意味着當你在最后的反向類中設置 self.head 時，你將頭部設置為非空白頭部，並在打印中跳過它。

為了彌補這一點，您需要創建一個新的空白頭並設置在 prev 旁邊：

    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

輸出是

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