```
# Definition for singly-linked list.
# class ListNode:
# def __init__(self, x):
# self.val = x
# self.next = None
# naive approach
# faster than 74.50% of Python3 online submissions
class Solution:
def removeNthFromEnd(self, head: ListNode, n: int) -> ListNode:
# first get all nodes
nodes = []
cur = head
while cur:
nodes.append(cur)
cur = cur.next
# see how many we got
tot = len(nodes)
start = cur = ListNode(0)
# construct nodes while skipping the Nth from the end
for i in range(tot):
if i != tot - n:
cur.next = nodes[i]
cur = cur.next
# clean up last node (only relevant if n == 1)
cur.next = None
return start.next
# "clever" approach
# https://leetcode.com/problems/remove-nth-node-from-end-of-list/discuss/8802/3-short-Python-solutions
class Solution:
def removeNthFromEnd(self, head, n):
fast = slow = head
# fast speed through the nodes to get to nth node
for _ in range(n):
fast = fast.next
# if fast is None, simply return head.next
# since we are skipping first element
if not fast:
return head.next
# now we step more carefully
# while fast.next exists (we haven't reached the end)
# we step through slow (which starts at head)
# and fast at the same time
while fast.next:
fast = fast.next
slow = slow.next
# when fast reached the last node
# slow is n-nodes away from the end,
# since fast had a n-node head start
# Thus, we skip the next node, and return the modified list of nodes
slow.next = slow.next.next
return head
```

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