從單鏈表到雙鏈表

Question

如何編寫像“next(lst)”這樣的函數來返回 PREVIOUS 值而不是 NEXT 值？

class EmptyNode():
    __slots__ = ()

class Node():
    __slots__ = ('data', 'next')


class MyList():
"""A class that encapsulates a node based linked list"""
    __slots__ = ('head', 'size', 'cursor')

def mkEmptyNode():
    return EmptyNode()

def mkNode(data, next):
    node = Node()
    node.data = data
    node.next = next
    return node

def mkMyList():
    lst = MyList()
    lst.head = mkEmptyNode()
    lst.size = 0
    lst.cursor = mkEmptyNode()
    return lst

在類似於['a','b','c']的鏈表中， next(lst)將返回'a' ，下一次返回'b' ，下一次返回'c' ，下次它會返回錯誤

def next(lst):
    if isinstance(lst.cursor, EmptyNode):
        raise IndexError("cursor is invalid")

    val = lst.cursor.data
    lst.cursor = lst.cursor.next
    return val

Answer 1

您將需要在每個列表項（ previous ）中維護一個額外的指針。

class EmptyNode():
  __slots__ = ()

class Node():
  __slots__ = ('data', 'next', 'prev')

class MyList():
  """A class that encapsulates a node based linked list"""
  __slots__ = ('head', 'size', 'cursor')

def mkEmptyNode():
  return EmptyNode()

def mkNode(prev, data, next):
  node = Node()
  node.prev = prev
  node.data = data
  node.next = next
  return node

def mkMyList():
  lst = MyList()
  lst.head = mkEmptyNode()
  lst.size = 0
  lst.cursor = mkEmptyNode()
  return lst

然后，您可以使用它向后導航列表：

def previous(lst):
  if isinstance(lst.cursor, EmptyNode):
    raise IndexError("cursor is invalid")

  val = lst.cursor.data
  lst.cursor = lst.cursor.prev
  return val

Answer 2

在常規雙向鏈表中，您將向 Node 類添加一個prev屬性（指向前一個節點）並添加一個 prev 方法，該方法執行如下操作：

class Node():
    __slots__ = ('data', 'next', 'prev')

進而：

def prev(lst):
    if isinstance(lst.cursor, EmptyNode):
        raise IndexError("cursor is invalid")

    val = lst.cursor.data
    lst.cursor = lst.cursor.prev
    return val