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

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.

Follow up:

Could you do this in one pass?

难度：medium

题目：

给定一链表，移除其倒数第n个结点。注意：n总是合法数

思路：双指针

Runtime: 6 ms, faster than 98.72% of Java online submissions for Remove Nth Node From End of List.

Memory Usage: 27 MB, less than 39.84% of Java online submissions for Remove Nth Node From End of List.

/** * Definition for singly-linked list. * public class ListNode { *     int val; *     ListNode next; *     ListNode(int x) { val = x; } * } */class Solution {    public ListNode removeNthFromEnd(ListNode head, int n) {        ListNode dummyHead = new ListNode(0);        dummyHead.next = head;        ListNode ptr = dummyHead,lastNPtr = dummyHead;        while (ptr.next != null) {            if (--n < 0) {                lastNPtr = lastNPtr.next;            }                        ptr = ptr.next;        }                lastNPtr.next = lastNPtr.next.next;                return dummyHead.next;    }}