删除链接列表中的最后一个节点

Question

我正在使用Python实现链表中最后一个节点的删除。以下是我的代码：

class Node:
def __init__(self, key):
self.key = key
self.next = None

def printList(head):
curr = head
while curr != None:
print(curr.key, end=" ")
curr = curr.next

def deleteLastNode(head):
if head == None:
return None

temp = head

# Iterating till the last Node
while temp.next != None:
temp = temp.next
temp = None
return head
# Driver code
head = Node(10)
head.next = Node(20)
head.next.next = Node(30)
head = deleteLastNode(head)
printList(head)

然而，在输出中，我仍然得到了完整的链表。

10 20 30

当最后一个节点已经设置为temp = None时，如何打印30？

Answer 1

好吧，你的链表是：

10 -> 20 -> 30
^head

您的迭代：

10 -> 20 -> 30
^head       ^temp

那么当你做temp = None时，就意味着(你只需将None分配给temp(：

10 -> 20 -> 30       None
^head                ^temp

正确的方法是在20上迭代时，执行temp.next = None以删除对最后一个节点的引用。所以你的代码可能是：

class Node:
def __init__(self, key):
self.key = key
self.next = None

def printList(head):
curr = head
while curr != None:
print(curr.key, end=" ")
curr = curr.next

def deleteLastNode(head):
if head == None:
return None

temp = head

# Iterating till the last Node
while temp.next.next != None:
temp = temp.next
temp.next = None
return head
# Driver code
head = Node(10)
head.next = Node(20)
head.next.next = Node(30)
head = deleteLastNode(head)
printList(head)

当您的链表至少包含两个元素时，此代码将起作用。当只有一个元素时，这将引发异常。我建议您使用next指向真实头部节点的虚拟头部节点。

Answer 2

#Another way of solving 
class Node:
def __init__(self, key):
self.key = key
self.next = None

def printList(head):
curr = head
while curr != None:
print(curr.key, end=" ")
curr = curr.next
def deleteLastNode(head):
if head == None:
return None
temp = head
prev=None    #creating the value of previous element
# Iterating till the last Node
while temp.next != None:
prev=temp  #updating for every iteration
temp = temp.next
prev.next = None #returning as NONE value
return head
# Driver code
head = Node(10)
head.next = Node(20)
head.next.next = Node(30)
head = deleteLastNode(head)
printList(head)