get(index)：获取链表中第 index 个节点的值。如果索引无效，则返回-1。
addAtIndex(index,val)：在链表中的第 index 个节点之前添加值为 val  的节点。如果 index 等于链表的长度，则该节点将附加到链表的末尾。如果 index 大于链表长度，则不会插入节点。
deleteAtIndex(index)：如果索引 index 有效，则删除链表中的第 index 个节点。

Design your implementation of the linked list. You can choose to use the singly linked list or the doubly linked list. A node in a singly linked list should have two attributes: val and next. val is the value of the current node, and next is a pointer/reference to the next node. If you want to use the doubly linked list, you will need one more attribute prev to indicate the previous node in the linked list. Assume all nodes in the linked list are 0-indexed.

get(index) : Get the value of the index-th node in the linked list. If the index is invalid, return -1.
addAtHead(val) : Add a node of value val before the first element of the linked list. After the insertion, the new node will be the first node of the linked list.
addAtTail(val) : Append a node of value val to the last element of the linked list.
addAtIndex(index, val) : Add a node of value val before the index-th node in the linked list. If index equals to the length of linked list, the node will be appended to the end of linked list. If index is greater than the length, the node will not be inserted.
deleteAtIndex(index) : Delete the index-th node in the linked list, if the index is valid.

Note:

All values will be in the range of [1, 1000].
The number of operations will be in the range of [1, 1000].

```public class Program {

public static void Main(string[] args) {

}

public class ListNode {
public int val;
public ListNode next;
public ListNode(int x) { val = x; }
}

private ListNode _listNode = null;
private int _count = 0;

_listNode = new ListNode(0);
}

public int Get(int index) {
if(index < 0 || index >= _count) return -1;
var idx = 0;
var node = _listNode;
while(idx++ != index) {
node = node.next;
}
return node.val;
}

if(_count == 0) {
_listNode.val = val;
} else {
}
_count++;
}

}

public void AddAtIndex(int index, int val) {
if(index < 0 || index > _count) return;
if(index == 0) {
_count++;
return;
}
var pre = _listNode;
var idx = 0;
while(idx++ != index - 1) {
pre = pre?.next;
}
if(pre == null) return;
var temp = pre.next;
pre.next = new ListNode(val);
pre.next.next = temp;
_count++;
}

public void DeleteAtIndex(int index) {
if(index < 0 || index >= _count) return;
var pre = _listNode;
var idx = 0;
while(idx++ != index - 1) {
pre = pre.next;
}
pre.next = pre.next.next;
_count--;
}
}

}```

```2
3```