Remove Nth Node From End of List

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?

SOLUTION:
# 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, n):
        slow = head
        fast = head
        for _ in range(n):
            fast = fast.next
        while fast and fast.next:
            slow = slow.next
            fast = fast.next
        if slow == head and not fast:
            return head.next
        slow.next = slow.next.next
        return head

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