链表--删除链表倒数第 n 个结点

给定一个链表，删除链表的倒数第 n 个节点，并且返回链表的头结点。

1->2->3->4->5 n = 2.
当删除了倒数第二个节点后，链表变为 1->2->3->5.

public static ListNode removeNthFromEnd(ListNode head, int n) {
        if (head == null) {
            return null;
        }

        ListNode node = head;
        //判断有多少个数据
        int k = 0;
        List<ListNode> list = new ArrayList<>();
        while (node != null) {
            k++;
            list.add(node);
            node = node.next;
        }
        //正数--需要删除的结点
        int removedNum = k + 1 - n;
        if (removedNum == k) {
            ListNode listNode = list.get(removedNum - 2);
            listNode.next = null;
        } else {
            ListNode listNode = list.get(removedNum - 1);
            ListNode node1 = listNode.next;
            listNode.next = node1.next;
            listNode.val = node1.val;
        }
        return head;
    }
 private static class ListNode {

        private int val;

        private ListNode next;

        public ListNode(int val) {
            this.val = val;
        }

        public ListNode(int val, ListNode next) {
            this.val = val;
            this.next = next;
        }
    }