如何在 python 中创建和清空链表节点？

Question

According to leetcode this is the definition of a linked list

# Definition for singly-linked list.
# class ListNode:
#     def __init__(self, val=0, next=None):
#         self.val = val
#         self.next = next

How do you create a single node that when called outputs

[]

instead of

[0]

In the merge two linked lists if we pass two empty nodes then the output is also an empty node, the question is how do you make an empty ListNode that has a value other than 0 or None

Answer 1

LeetCode lets you specify linked lists using Python list notation for executing tests in the browser. This is not how the linked list is actually represented in memory; LeetCode runs code on their backend to take the list you input as a test case and turn it into a linked list.

If you're trying to test your code locally and want to represent an empty linked list, use None rather than [] . A linked list with one node can be declared as follows:

singleton_linked_list = ListNode(5) # Creates a new linked list with one node.

Answer 2

If you want to create Empty Linked List and assuming head is given.

curr = head
curr.next = None
curr = curr.next

Answer 3

You can use the following while solving your problem in LeetCode:

if not lists:
   return None
if len(lists) == 1:
   return lists[0]

Thus, whenever there is input like [] or [[]] , so after execution LeetCode compiler will return an empty List with None.

Answer 4

I had the same problem as you but I just figured it out. All you have to do is return None and Leetcode will accept it. I tested it on Leetcode problem 23, Merge k Sorted Lists.