Leetcode remove Nth node from End of List removes the last n nodes

Topic:

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

For example,

   1->2->3->4->5  N = 2.   1->2->3->5.

Note:
Given n would always be valid.
Try to do the in one pass.

Translation:

Give you a list of the last nth nodes to remove.

Ideas:

The difficulty of this problem is also OK, is some of the details of the aspect, the first possible link list on a node. The second possibility is that the first node is deleted. The last is the connection to the node 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 used the method is two pointer p,q,q first traverse to the position of n-1, and then the two pointers traverse backwards at the same time, until Q to the end, at this time p should be deleted. At the same time, a pointer points to the previous one of P.

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

If P and pre are not equal, the P is deleted directly

Leetcode remove Nth node from End of List removes the last n nodes