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Today, I would like to share with you how Java deletes the penultimate node of the linked list. The content is detailed and the logic is clear. I believe most people still know too much about this knowledge, so share this article for your reference. I hope you can get something after reading this article.

Problem-solving ideas

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 first, and the back pointer is second, both equal to pre

First moves forward n steps first.

Then first and second move forward together, where the distance between them is n, and when the first is to the tail, the position of the second happens to be the last node of the penultimate node.

Java code class Solution {public ListNode removeNthFromEnd (ListNode head, int n) {ListNode pre = new ListNode (0); pre.next = head; ListNode first= pre; ListNode second = pre; while (n > 0) {first=first.next; NMurray;} while (first.next! = null) {first=first.next Second=second.next;} second.next = second.next.next; return pre.next;}} above is all the content of the article "how to delete the penultimate node of the linked list by Java". Thank you for reading! I believe you will gain a lot after reading this article. The editor will update different knowledge for you every day. If you want to learn more knowledge, please pay attention to the industry information channel.

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