我在尾部函数中的插入节点不会改变尾部的值

Question

所以我试图编写一个函数，它将链表的头部和数据作为参数，并在链表的尾部附加一个节点。当它从函数内部打印“新”头时，它看起来工作正常（请注意，链表的头最初不是节点，它实际上等于 None）（head 成为 head.data = 0 and head.next = None)，但是当我在调用函数后打印链表的头部时它不返回节点，实际上它返回None，这意味着函数没有在尾部插入节点最初是头部。有人可以解释为什么头部的价值没有改变吗？

class Node:

    def __init__(self, data):
        self.data = data
        self.next = None

class LinkedList:

    def __init__(self):
        self.head = None

def InsertNode(head, data):

    if head == None:
        print("head == None, so head will be a node now")
        head = Node(data)
        print("now, head is a node where head.data = {} and head.next = {}".format(head.data, head.next))

    else: 
        tail = head
        while tail.next != None:
            tail = tail.next
            tail.next = Node(data)

def PrintLL(linkedList):

    node = linkedList.head

    while node != None:
        print (node.data)

    node = node.next

llist = LinkedList()
##print("llist.head == {}".format(llist.head))

InsertNode(llist.head, 0)
print("after InsertNode with data = 0, llist.head == {}".format(llist.head))

Answer 1

以下是固定代码：

class Node:

    def __init__(self, data):
        self.data = data
        self.next = None

class LinkedList:

    def __init__(self):
        self.head = None

def InsertNode(node, data):

    if node.head == None:
        print("head == None, so head will be a node now")
        node.head = Node(data)
        print("now, head is a node where head.data = {} and head.next = {}".format(node.head.data, node.head.next))

    else: 
        tail = node.head
        while tail.next != None:
            tail = tail.next
            tail.next = Node(data)

def PrintLL(linkedList):

    node = linkedList.head

    while node != None:
        print (node.data)

    node = node.next

llist = LinkedList()
##print("llist.head == {}".format(llist.head))

InsertNode(llist, 0)
print("after InsertNode with data = 0, llist.head == {}".format(llist.head))

发生的事情是你传递了一个指向头部的指针：

InsertNode(llist.head, 0)

InsertNode 然后会为该变量分配一个新的节点实例：

head = Node(data)

由于头指针是通过赋值传递的，函数InsertNode本地的head的版本发生了变化，但原来的llist.head不受影响。

Answer 2

您需要将LinkedList 对象本身传递给InsertNode，而不是其head 属性。通过将head 属性传递给InsertNode，对其进行的任何新分配都不会反映在原始对象上。

class Node:

    def __init__(self, data):
        self.data = data
        self.next = None

class LinkedList:

    def __init__(self):
        self.head = None

def InsertNode(linkedList, data):

    if linkedList.head == None:
        print("head == None, so head will be a node now")
        linkedList.head = Node(data)
        print("now, head is a node where head.data = {} and head.next = {}".format(linkedList.head.data, linkedList.head.next))

    else:
        tail = linkedList.head
        while tail.next != None:
            tail = tail.next
            tail.next = Node(data)

def PrintLL(linkedList):

    node = linkedList.head

    while node != None:
        print (node.data)

    node = node.next

llist = LinkedList()

InsertNode(llist, 0)
print("after InsertNode with data = 0, llist.head == {}".format(llist.head.data))