将链表基本复制到另一个LinkedLis

Question

我不想做一个长时间的电脑编程。我今年25岁，正在学习这些概念和python，所以请友好一点。在

我想使用一个名为copyList的函数将LinkedList从一个LinkedList对象复制到另一个对象。它不应该接受除自身之外的任何参数，并且应该在不改变原始列表的同时输出LinkedList的副本。在

我试着查看堆栈，发现了一个类似的代码，但它并没有解决我的问题，因为代码与我的代码相似，但不起作用，因为我尝试打印新的LinkedList，它不包含值，我相信它是空的。我在下面给出了类似的代码：

class Node:
    def __init__(self,x):
        self.data = x
        self.next = None
class LinkedList:
    def __init__(self):
        self.top = None
    def printList(self):
        print("top^")
        while self.top is not None:
            print(self.top.data)
            self.top = self.top.next
        print("tail^")
    def in_list(self, x):
        current = self.top 
        while current is not None:
            if current.data is x:
                return 1
            current = current.next
        return 0
    def findCellBefore(self, x):
        current = self.top
        if current is None:
            return 0
        while current.next is not None:
            if current.next.data is x:
                current = current.next
        return 1
    def findCellBeforeSential(self,x):
        if (self.top.next is None):
            return 0 
        while (self.top.next is not None):
            if (self.top.next.data is x):
                return self.top.data
        return 1
    def add_0(self, newNode):
        # i. make next of new node as head.
        newNode.next = self.top
        # ii. move head to point to new node.
        self.top = newNode
    def add_end(self, newNode):
        current = self.top
        if (current is None):
            self.add_0(newNode)
            return 
        while (current.next is not None):
            current = current.next

        current.next = newNode 
        newNode.next = None
    def insertNode(self, after_me, new_cell):
        new_cell.next = after_me.next 
        after_me.next = new_cell

        # update prev links.

        new_cell.next.prev = new_cell 
        new_cell.prev = after_me 
    def  deleteAfter(self, after_me):
            after_me.next = after_me.next.next

    def CopyList(self):#Out put new Linked List that is a copy of current Linked List with out altering it. 
        # create new LinkedList
        newLinkedList = LinkedList()
        #current = self.top
        #below is from stackoverflow : https://stackoverflow.com/questions/36491307/how-to-copy-linked-list-in-python
        #while current.next != None:
        #    newLinkedList.add_end(current.data)
        #    current = current.next
        #newLinkedList.add_end(current.data)
        #return newLinkedList
        while self.top is not None:
            newNode = Node(self.top.data)
            newLinkedList.add_end(newNode)
            self.top = self.top.next
        return newLinkedList

LIST0 = LinkedList()

node0 = Node(1)
node1 = Node(2)
node2 = Node(3)
LIST0.add_end(node1)
LIST0.add_0(node0)
LIST0.add_0(node2)
node3 = Node(4)
LIST0.insertNode(node2, node3)

LIST0.printList()

LIST1=LIST0.CopyList()

LIST1.printList()

我想让它简单地打印出一个新列表，它是LIST0的副本，并且让LIST1作为LinkedList对象工作。在

Answer 1

我认为这与newLinkedList的创建没有被保存在内存中有关，因为代表您的某种指针错误（也就是说，您需要进一步学习Python（nic）语言）。正在传递，但新节点未保存到newLinkedList中。我也希望看到解决办法。在

Answer 2

问题的很大一部分是：

while self.top is not None:
    newNode = Node(self.top.data)
    newLinkedList.add_end(newNode)
    self.top = self.top.next

正常情况下self.top公司指向节点的顶部，除非替换或删除顶部节点，否则不应更改。你要做的基本上就是从列表中删除所有节点。在

注释掉的代码看起来是正确的，除了以下行“newLinkedList.add_结束(当前数据)“不够缩进。Python缩进系统的一个抱怨是，如果代码以改变缩进的方式粘贴，也会改变行的分组。该行应该是循环的一部分，并与上面的行的缩进相匹配。在

Answer 3

看完你的代码后，@johnbayko似乎是正确的，你必须离开self.top公司（哨兵头）独自一人。。。在您的代码中，您似乎有某些函数使用self.top公司out right和其他使用current=self.top公司引用它并在函数内完成相应的工作。函数的顶部指针应该正好是这样的，因为它应该是一个参考（普遍-认为它是“北极星”，供您在导航列表时使用），以供其余代码遵循。在

下面是更正的代码：在此之后理解链表等概念应该更容易些。在

class Node:
    def __init__(self,x):
        self.data = x
        self.next = None
class LinkedList:
    def __init__(self):
        self.top = None
    def printList(self):
        print("top^")
        current = self.top
        while current is not None:
            print(current.data)
            current = current.next
        print("tail^")
    def in_list(self, x):
        current = self.top 
        while current is not None:
            if current.data is x:
                return 1
            current = current.next
        return 0
    def findCellBefore(self, x):
        current = self.top
        if current is None:
            return 0
        while current.next is not None:
            if current.next.data is x:
                current = current.next
        return 1
    def findCellBeforeSential(self,x):
        current = self.top
        if (current.next is None):
            return 0 
        while (current.next is not None):
            if (current.next.data is x):
                return current.data
        return 1
    def add_0(self, newNode):
        # i. make next of new node as head.
        newNode.next = self.top
        # ii. move head to point to new node.
        self.top = newNode
    def add_end(self, newNode):
        current = self.top
        if (current is None):
            self.add_0(newNode)
            return 
        while (current.next is not None):
            current = current.next

        current.next = newNode 
        newNode.next = None
    def insertNode(self, after_me, new_cell):
        new_cell.next = after_me.next 
        after_me.next = new_cell

        # update prev links.

        new_cell.next.prev = new_cell 
        new_cell.prev = after_me 
    def  deleteAfter(self, after_me):
            after_me.next = after_me.next.next

    def CopyList(self):#Out put new Linked List that is a copy of current Linked List with out altering it. 
        # create new LinkedList
        newLinkedList = LinkedList()
        current = self.top
        #below is from stackoverflow : https://stackoverflow.com/questions/36491307/how-to-copy-linked-list-in-python
        while current is not None:
            newNode = Node(current.data)
            newLinkedList.add_end(newNode)
            current = current.next
        return newLinkedList
        print("here")

        #current = self.top
        #print(current)
        #while current.next is not None:
        #    print(0)
        #    newNode = Node(self.top.data)
        #    print(1)
        #    newLinkedList.add_end(newNode)
        #    print(2)
        #    self.top = self.top.next
        return newLinkedList

LIST0 = LinkedList()

node0 = Node(1)
node1 = Node(2)
node2 = Node(3)
LIST0.add_end(node1)
LIST0.add_0(node0)
LIST0.add_0(node2)
node3 = Node(4)
LIST0.insertNode(node2, node3)

LIST0.printList()

LIST1=LIST0.CopyList()

LIST1.printList()