使用 Python 实现双向链表

Question

我正在尝试在 python 中实现双向链表，但无法正确实现删除。

def deleteFromBeginning(self):
    if self.listLength()==0:
        print("Can't delete from an empty list")
    else:
        self.head = self.head.next

def deleteFromEnd(self):
    if self.listLength() ==0:
        print("Can't delete from an empty list")
    else:

        current=self.head
        while(current.next!=None):
            current.prev = current
            current = current.next

        current.prev.next = None
        current.prev = None

def deleteFromPosition(self,pos):
    if pos> self.listLength() or pos<0:
        ValueError("Enter a valid position")
    elif pos==0:
        deleteFromBeginning()
    elif pos == self.listLength():
        deleteFromEnd()
    else:
        if self.listLength()==0:
            print("Can't delete from an empty list")
        else:
            current=self.head
            count = 1
            while(current.next!=None and count<pos):
                current = current.next
                count = count+1

            current.prev.setNext(current.next)
            current.next.setPrev(current.prev)
            current.setPrev(None)
            current.setNext(None)

def deleteWithData(self,data):
    if self.listLength()==0:
        ValueError("Can't delete from an empty list")
    else:
        current = self.head
        while current.next!=None:
            if current.data == data:
                current.prev.next = current.next
                current.next.prev = current.prev
                return
            else:
                current = current.next
        print("The value provided is not present")

从头到尾的删除工作正常，但删除位置和数据并没有给出预期的结果。

Answer 1

我会做的一个轻微的重构是移动你的

if self.listLength()==0:
            print("Can't delete from an empty list")

检查到最顶部，而不是在 else 内部。 这应该是您首先进行的检查之一。

除此之外，您可以尝试跟踪前一个节点：

while(current.next!=None and count<pos):
                temp = current
                current = current.next
                count = count+1

        temp.setNext(current.next)
        current.next.setPrev(temp)

这应该通过将前一个节点重新链接到下一个节点来删除 x 位置的节点。

还没有测试过，让我知道它是怎么回事！