删除链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。 下面是我的代码：

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)

但是，在输出中我仍然得到完整的链表。

10 20 30

我的疑问是为什么最后一个节点没有被删除。 此外，当最后一个节点已设置为temp = None时，它如何打印值30 ？

Answer 1

好吧，你的链表是：

10 -> 20 -> 30

^head

你的迭代：

10 -> 20 -> 30

^head       ^temp

然后当你做temp = None ，只是意味着（你只是将None分配给temp ）：

10 -> 20 -> 30       None

^head                ^temp

正确的方法是当您迭代20 ，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)

当您的链表包含至少两个元素时，此代码将起作用。 当只有一个元素时，这将引发异常。 我建议您使用一个虚拟头节点，它next指向真正的头节点。

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)