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Editor to share with you how to delete the penultimate node of the linked list in leetcode, I believe most people do not know much about it, so share this article for your reference. I hope you will gain a lot after reading this article. Let's learn about it together.

Topic description

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?

The idea of solving the problem

Tags: linked list

The overall idea is to let the front pointer move n steps first, and then move the front and rear pointers together until the front pointer reaches the tail.

First of all, set up the advance pointer pre, which is a little trick, which is explained in question 2.

Let the next node of the pre-pointer pre point to head, the front pointer is start, and the back pointer is end, both equal to pre

Then start and end move forward together, and the distance between them is n. When start reaches the tail, the position of end happens to be the last but one node.

Because the node is to be deleted, it cannot be deleted until it is moved to the previous node, so the loop end condition is start.next! = null

Return pre.next after deletion, why not return head directly, because head may be the deleted point

Time complexity: O (n)

Code / * Definition for singly-linked list. * public class ListNode {* int val; * ListNode next; * ListNode (int x) {val = x;} *} * / class Solution {public ListNode removeNthFromEnd (ListNode head, int n) {ListNode pre = new ListNode (0); pre.next = head; ListNode start = pre, end = pre; while (n! = 0) {start = start.next While (start.next! = null) {start = start.next; end = end.next;} end.next = end.next.next; return pre.next;}}

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