# Remove Nth node from the end of a linked list

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

```
class Solution:
def removeNthFromEnd(self, head: ListNode, n: int) -> ListNode:
fast, slow = head, head
for _ in range(n): fast = fast.next
if not fast: return head.next
while fast.next: fast, slow = fast.next, slow.next
slow.next = slow.next.next
return head
```

## Related Problems

Given the head of a singly linked list, reverse the list, and return *the reversed list*.

You are given the heads of two sorted linked lists list1 and list2.

Merge the two lists into one **sorted** list. The list should be made by splicing together the nodes of the first two lists.

Return *the head of the merged linked list*.

Given the root of a binary tree, flatten the tree into a "linked list":

The "linked list" should use the same TreeNode class where the right child pointer points to the next node in the list and the left child pointer is always null.

The "linked list" should be in the same order as a pre-order traversal of the binary tree.

Given the heads of two singly linked-lists headA and headB, return *the node at which the two lists intersect*. If the two linked lists have no intersection at all, return null.