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?

Solution for this problem statement:

```
public ListNode removeNthFromEnd(ListNode head, int n) {
ListNode dummy = new ListNode(0);
dummy.next = head;
ListNode first = dummy;
ListNode second = dummy;
// Advances first pointer so that the gap between first and second is n nodes apart
for (int i = 1; i <= n + 1; i++) {
first = first.next;
}
// Move first to the end, maintaining the gap
while (first != null) {
first = first.next;
second = second.next;
}
second.next = second.next.next;
return dummy.next;
}
```

There is another solution is available for this problem statement:

```
public ListNode removeNthFromEnd(ListNode head, int n) {
ListNode dummy = new ListNode(0);
dummy.next = head;
int length = 0;
ListNode first = head;
while (first != null) {
length++;
first = first.next;
}
length -= n;
first = dummy;
while (length > 0) {
length--;
first = first.next;
}
first.next = first.next.next;
return dummy.next;
}
```

There are other articles are available which give you detailed information about the array. Also, there are some linked list-related programs are available like Add two numbers using a linked list.

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