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This article Xiaobian introduces in detail for you "python how to delete the penultimate node of the linked list", the content is detailed, the steps are clear, and the details are handled properly. I hope that this article "python how to delete the penultimate node of the linked list" can help you solve your doubts.

[title]

Given a linked list, delete the penultimate node of the linked list and return the header node of the linked list.

Example:

Given a linked list: 1-> 2-> 3-> 4-> 5, and n = 2.

When the penultimate node is deleted, the linked list becomes 1-> 2-> 3-> 5.

Description:

The given n guarantee is valid.

Advanced:

Can you try to do it with a scan?

[ideas]

Solution 1: traversing the linked list to get the length of the linked list N. Then delete the last-to-last node, that is, delete the Nth node. Find the N-n-1 node and write it down as pmaine Q = p.next, p.next=p.next.next, del p (remember to clear memory).

The only question is: how to delete the first element, which needs to be judged separately? You don't have to go to so much trouble: add a meaningless header node, and all the deletion logic becomes consistent!

Solution 2: use two pointers first and second to traverse the linked list. First, the first pointer moves forward n steps, and the second pointer remains the same; then, the first pointer and the second pointer advance at the same time until first.next is None. At this point, the second pointer points to the penultimate node, and second.next = second.next.next can delete the useless memory at the same time. (the code logic of deleting the first element is different from that of other elements, and the solution refers to the description of solution 1. )

[code]

Python version

# Definition for singly-linked list.

# class ListNode:

# def _ _ init__ (self, val=0, next=None):

# self.val = val

# self.next = next

Class Solution:

Def removeNthFromEnd (self, head: ListNode, n: int)-> ListNode:

# add empty header nodes to make logic consistent

Node = ListNode (val=0, next=head)

Head = node

# find Node n-1

Count = 0

P = head

While count < n:

Count + = 1

P = p.next

# find the last-to-last node

Q = head

While p.next:

P = p.next

Q = q.next

# Delete the penultimate node

R = q.next

Q.next = q.next.next

Del r

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