Remove nth node from end of list

# Definition for singly-linked list.
# class ListNode:
#     def __init__(self, val=0, next=None):
#         self.val = val
#         self.next = next
class Solution:
    def removeNthFromEnd(self, head: Optional[ListNode], n: int) -> Optional[ListNode]:
        dummy = ListNode()
        dummy.next = head
        curr = dummy
        prev = dummy

        for i in range(n + 1):
            curr = curr.next

        while curr:
            curr = curr.next
            prev = prev.next

        prev.next = prev.next.next

        return dummy.next

Remove Nth Node From End of List

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

 

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

 

Follow up: Could you do this in one pass?