通过递归方式反向双向链表不在python中迭代答案

【问题标题】：reverse doubly linked list by recursive way not iterate in python通过递归方式反向双向链表不在python中迭代
【发布时间】：2019-08-09 10:33:42
【问题描述】：

我如何通过递归为双链表编写反向函数。我已经在python中使用递归和重写来引用问题反向双链表，但它让我陷入无限循环，所以我重写了逻辑，但我有点失去了前一个点

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()

【问题讨论】：

通过将 dll.reverse() 更改为 dll.reverse_recursive() 我似乎可以正常工作。列表向后打印，头部指向带有val=4 的元素。您能否描述一下您的代码所做的更多不需要的内容？
谢谢你反向反向是我的错字。在self.head = head 之后，实际上我认为我只是通过重写反向函数的逻辑来修复我的代码。我会删除帖子

标签： python-3.x doubly-linked-list

【解决方案1】：

我所做的只是在最后添加一些打印输出，看看在哪里。在我看来，您的代码就像您期望的那样。在reverse() 函数之后，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)

输出：

Head:  1
1
2
3
4

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

【讨论】：

酷，谢谢。我只是更新我的原始帖子。所以我的原始代码无论如何都不起作用。

【解决方案2】：

我将在这里发布我的重写反向逻辑，它现在似乎有效。免费评论吧

    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)

【讨论】：