Remove Nth Node From End of List

1. Problem

 去掉链表的倒数第n个节点，并返回链表头。一次遍历完成

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.

2. Solution

用两个指针：

• p：指向可能的待删节点的前一个节点；
• q：q-p = n；

为了方便处理，在链表前加上一个空节点。代码如下：

/**
 * Definition for singly-linked list.
 * public class ListNode {
 *     int val;
 *     ListNode next;
 *     ListNode(int x) { val = x; }
 * }
 */
public class Solution {
    public ListNode removeNthFromEnd(ListNode head, int n) {
        if( head == null )
            return head;
        //create a new head node so that we can always remove p.next
        ListNode newHead = new ListNode(0);
        newHead.next = head;
        ListNode p = newHead;   //point to the previous node before to-be-deleted node
        ListNode q = head;  //q - p = n;
        int count = 1;  //to find the first satisfied q;
        for( ; q.next!=null && count<n; count++, q = q.next);
        //the to-be-deleted node exists
        if(count == n ){
            while( q.next != null ){
                p = p.next;
                q = q.next;
            }
            p.next = p.next.next;
        }
        return newHead.next;
    }
}