LeetCode 19 Remove Nth Node From End of List Remove the Nth Node, leetcodenth

LeetCode 19 Remove Nth Node From End of List Remove the Nth Node, leetcodenth

Question:

Given a linked list, remove the nth node from the end of list and return its head.

For example,

   Given linked list: 1->2->3->4->5, and n = 2.   After removing the second node from the end, the linked list becomes 1->2->3->5.

Note:
Given n will always be valid.
Try to do this in one pass.

Translation:

Give you a linked list and remove the nth last node.

Ideas:

The difficulty of this question is okay, that is, some details. First, the linked list may be a node. The first node may be deleted. The last step is to delete the node connection after the node is deleted.

Code:

 public ListNode removeNthFromEnd(ListNode head, int n) {         if(head==null||(head.next == null && n ==1))        return null;                ListNode p = head;        ListNode q = head;        ListNode pre = head;        while(n!=1)        {        q = q.next;        n--;        }        while(q.next!=null)        {            pre = p;        q = q.next;        p = p.next;        }        if(pre.next == p.next)            head = head.next;        else             pre.next = p.next;        return head;            }

I am using two pointers, p, q, and q, first traverse to the n-1 position, and then both pointers traverse backward at the same time until q ends. At this time, p should be deleted. At the same time, a pointer points to the first one of p.

If p and pre point to the same node, the first element is deleted. Because the previous q. next = null, this is as long as the head node points to his next node.

If p and pre are different, delete p directly.