排序双链表Python

Question

我在理解和实现双重链接列表时遇到了困难。我能掌握链表的大部分概念。以下是我目前为止的代码（用Python编写）

*这是纯粹的学术练习。我通常使用list和dict

class DoublyNode(object):
    """A node of the SortedDoublyLL object.

    DoublyNode(item, next=None, previous=None) -> a new DoublyNode with data as
    its data, and next and previous as its neighbors."""

    def __init__(self, data, next = None, previous = None):
        """Make a new DoublyNode from item, pointing to next and previous."""

        self.data = data
        self.next = next
        self.previous = previous

class SortedDoublyLL(object):
    """A Sorted Doubly Linked List.

    SortedDoublyLL() -> new SortedDoublyLL list that is empty
    SortedDoublyLL(sequence) -> a SortedDoublyLL initialized from sequence's
    items.

    """

    def __init__(self, sequence = []):
        """Make a new SortedDoublyLL from the elements of sequence."""

        if len(sequence) == 0:
            self.head = None
            self.tail = None
        else:
            cur_node = None
            prev_node = None
            sequence.sort()
            sequence.reverse()
            for element in sequence:
                prev_node = cur_node
                cur_node = DoublyNode(element, cur_node, prev_node)

            self.head = cur_node
            self.tail = DoublyNode(sequence[0])

Answer 1

将循环改为

for element in sequence:
    prev_node = cur_node
    cur_node = DoublyNode(element, None, prev_node)
    prev_node.next = cur_node

因为第prev_node = cur_node行在调用DoublyNode(element, cur_node, prev_node)之前，所以您最终将上一个元素和下一个元素都设置为上一个元素，这样您就得到了一个只有两个指向上一个元素的链接的链表。所以您也可以只传递None作为next参数¹，然后在下一次循环中手动初始化它。这样做的好处是在列表的最后一个元素上保留为None。在

1在构造函数中使用名称next作为参数将隐藏内置函数next，该函数推进迭代器。您可以使用名称next_，这是规范的做法。使用next作为属性不是问题，因为这样可以限定名称，这样就不会出现阴影。不过，它会在一些语法高亮中出错。在