我正在尝试创建一个函数，该函数接收链表和值的头部，创建具有该值的节点，将node.next设置为head，然后将新节点更新为head。我得到了它的工作，但我觉得我通过返回值非常低效地做这个，所以我想知道是否有更优雅的方法来做到这一点。
这是我正在使用的代码：
#!/usr/bin/env python3
"""
This module is a linked list implementation.
"""
class Node:
"""
Node structure to be used in our linked list.
"""
def __init__(self, value):
self.value = value
self.next = None
def linked_insert(head, value):
"""
Insert new node to the tail of the linked list.
Time Complexity: O(n)
"""
current = head
while current.next is not None:
current = current.next
current.next = Node(value)
def linked_insert2(head, value):
"""
Insert new node to the head of the linked list, making sure to update head.
Time Complexity: O(1)
"""
to_insert = Node(value)
to_insert.next = head
head = to_insert
return head
def linked_extract(head):
"""
Extract the last element in the linked list
"""
current = head
while current.next.next is not None:
current = current.next
tail = current.next
current.next = None
return tail
def linked_display(head):
"""
Print all node values in the linked list
"""
current = head
while current is not None:
print(current.value)
current = current.next
# Test Program
head = Node(5)
# linked_insert(head, 1)
# linked_insert(head, 2)
# linked_insert(head, 3)
#
# print(linked_extract(head).value)
head = linked_insert2(head, 1)
head = linked_insert2(head, 2)
head = linked_insert2(head, 3)
linked_display(head)
有没有办法在不必返回head的值并在测试程序中设置head =返回值的情况下执行此操作？
有问题的函数是linked_insert2()。整个程序是python中的链表实现。
使用链表类实现：
#!/usr/bin/env python3
"""
This module is a linked list implementation.
"""
class Node:
"""
Node class to be used in our linked list.
"""
def __init__(self, value):
self.value = value
self.next = None
class Linkedlist:
"""
Linked list implementation that supports functions including inserting
at head, inserting at tail, extracting elements, and displaying all
elements in the ist
"""
def __init__(self, head):
self.head = Node(head)
def insert(self, value):
"""
Insert new node to the tail of the linked list.
Time Complexity: O(n)
"""
current = self.head
while current.next is not None:
current = current.next
current.next = Node(value)
def insert2(self, value):
"""
Insert new node to the head of the linked list, making sure to update head.
Time Complexity: O(1)
"""
to_insert = Node(value)
to_insert.next = self.head
self.head = to_insert
def extract(self):
"""
Extract the last element in the linked list
"""
current = self.head
while current.next.next is not None:
current = current.next
tail = current.next
current.next = None
return tail
def display(self):
"""
Print all node values in the linked list
"""
current = self.head
while current is not None:
print(current.value)
current = current.next
# Test Program
head = Linkedlist(5)
# linked_insert(head, 1)
# linked_insert(head, 2)
# linked_insert(head, 3)
head.insert2(1)
head.insert2(2)
head.insert2(3)
head.display()