leetcode Remove Nth Node From End of List

题目：

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.

解题思路：

利用双指针的思想，先让前指针移动n步，然后前后指针一起移动。当前指针移动到链表结尾时，后指针恰好指在索要删除的节点。

代码：

 public ListNode removeNthFromEnd(ListNode head, int n) {        if (head == null || head.next == null)            return null;        ListNode first = head;        ListNode second = head;        for (int i = 0; i < n; i++) {            first = first.next;        }        if (first == null)            return head.next;        while (first.next != null) {            first = first.next;            second = second.next;        }        second.next = second.next.next;        return head;    }