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

Example 3:

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

Constraints:

The number of nodes in the list is sz.

1 <= sz <= 30

0 <= Node.val <= 100

1 <= n <= sz

Solution:

Approach 1:

```
class Solution {
public ListNode removeNthFromEnd(ListNode head, int n)
{
int length=0;
ListNode dummy=head;
while(dummy!=null)
{
length++;
dummy=dummy.next;
}
if(n>length) return dummy;
if(n==length) return head.next;
length-=n;
dummy=head;
while(length>1 && dummy.next!=null)
{
dummy=dummy.next;
length--;
}
dummy.next=dummy.next.next;
return head;
}
}
```

Approach 2:

```
class Solution {
public ListNode removeNthFromEnd(ListNode head, int n) {
ListNode dummy = new ListNode(0);
dummy.next = head;
ListNode fast = dummy;
ListNode slow = dummy;
for (int i = 1; i <= n + 1; i++) {
fast = fast.next;
}
while (fast != null) {
fast = fast.next;
slow = slow.next;
}
slow.next = slow.next.next;
return dummy.next;
}
}
```

## Comments