通过递归方式反向双向链表不在python中迭代

Question

我如何通过递归双链表来编写反向函数。我在python中使用递归和重写来引用反向双链表的问题，但是它带给我无限循环所以我重写了逻辑但是我有点失去了上一点

class Node:
    def __init__(self, data, prev=None, nxt=None):
        self.val = data
        self.prev = prev
        self.next = nxt


class DoublyLinkedList:
    def __init__(self, head):
        self.head = head

    def print_list(self):
        cur = self.head
        while cur is not None:
            print(cur.val)
            cur = cur.next

    def reverse(self):
        if self.head is None or self.head.next is None: return self.head
        cur = self.head
        def reverse_node(node):
            if node is None: return
            if node.next is None:
                node.prev = None
                return node
            new_head = reverse_node(node.next)
            new_node = node.next
            tmp = new_node.next
            new_node.prev = tmp
            new_node.next = node
            node.next = None
            return new_head
        self.head = reverse_node(cur)

a = Node(1, prev=None)
b = Node(2, prev=a)
c = Node(3, prev=b)
d = Node(4, prev=c)
a.next = b
b.next = c
c.next = d
dll = DoublyLinkedList(a)
dll.print_list()
dll.reverse()
dll.print_list()

Answer 1

我所做的就是在最后添加一些打印输出以查看其中的内容。它看起来好像你的代码符合你的期望。在reverse()功能之后，头部似乎清楚地指向d而不是a

class Node:
    def __init__(self, data, prev=None, nxt=None):
        self.val = data
        self.prev = prev
        self.next = nxt


class DoublyLinkedList:
    def __init__(self, head):
        self.head = head

    def print_list(self):
        cur = self.head
        while cur is not None:
            print(cur.val)
            cur = cur.next

    def reverse(self):
        if self.head is None or self.head.next is None: return
        cur = self.head

        def reverse_node(node):
            if node is None: return node
            node.next, node.prev = node.prev, node.next
            if node.prev is None: return node
            return reverse_node(node.prev)

        self.head = reverse_node(cur)

a = Node(1, prev=None)
b = Node(2, prev=a)
c = Node(3, prev=b)
d = Node(4, prev=c)
a.next = b
b.next = c
c.next = d
dll = DoublyLinkedList(a)
print("Head: ",dll.head.val)
dll.print_list()
dll.reverse()
print()
print("Head: ",dll.head.val)
dll.print_list()
print("Is the head at a? ",dll.head is a)
print("Is the head at d? ",dll.head is d)

OUTPUT：

Head:  1
1
2
3
4

Head:  4
4
3
2
1
Is the head at a?  False
Is the head at d?  True

Answer 2

我将在这里发布我的重写反向逻辑，它现在似乎有效。免费评论它

    def reverse(self):
        if self.head is None or self.head.next is None: return
        cur = self.head

        def reverse_node(node):
            if node is None: return node
            node.next, node.prev = node.prev, node.next
            if node.prev is None: return node
            return reverse_node(node.prev)
        self.head = reverse_node(cur)