Leetcode-Remove Nth Node From End of List

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.

分析是：1.逆序链表，时间开销貌似较大，而且也很麻烦，和扫描两次链表没啥区别，不做

           2. 观察，如果放两个指针，p 和q，如果q指向链表尾，p指向第n个节点的话（反向数节点），那么刚好，也就是说保证p和q指针的距离，扫描链表，等q到链表尾了，p也就是指向了想要remove的节点了

  特殊情况就是如果要删的是head节点（原因在于p，q初始都从head开始）

  另外要注意的是删节点需要知道前驱，所以遍历到待删除节点的前驱就不要再扫描了

        public ListNode removeNthFromEnd(ListNode head, int n) {
             ListNode p = head;
             ListNode q = head;
             
             //Init to make sure the distance between p and q is (1,n)
             int counter  = 1;
             while(counter < n){
                q  = q.next;
                counter ++;
             }
             
             //if n points to the head
             if(q.next == null){
                 return head.next;
             }
             
             //remove a node needs to know its previous node
             while(q.next.next !=null){
                 p = p.next;
                 q = q.next;
             }
             //remove the node 
              p.next = p.next.next;
             
             return head;
        
        }