Given the `head` of a linked list, remove the `nth` node from the end of the list and return its head.

Follow up: Could you do this in one pass?

Example 1:

```Input: head = [1,2,3,4,5], n = 2
Output: [1,2,3,5]
```

Example 2:

```Input: head = , n = 1
Output: []
```

Example 3:

```Input: head = [1,2], n = 1
Output: 
```

Constraints:

• The number of nodes in the list is `sz`.
• `1 <= sz <= 30`
• `0 <= Node.val <= 100`
• `1 <= n <= sz`
``````/**
* function ListNode(val, next) {
*     this.val = (val===undefined ? 0 : val)
*     this.next = (next===undefined ? null : next)
* }
*/
/**
* @param {number} n
* @return {ListNode}
*/
var removeNthFromEnd = function(head, n) {
return null;

let first = new ListNode();
let second = first;
// first point to n + 1
first = first.next;
while (n > 0) {
first = first.next;
n--;
}
// first already reach to the end null, means need to remove the first node
if (first === null) {
} else {
// maintaining N nodes apart between first and second pointer
while (first !== null) {
first = first.next;
second = second.next;
}
// Now second pointer to the node that need to be removed
// remove that node now
second.next = second.next.next;
}