# LeetCode 019Remove Nth Node From End of List

## LeetCode 019 Remove Nth Node From End of List Problem

``````# 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.
#
# Could you do this in one pass?

# class ListNode(object):
#     def __init__(self, x):
#         self.val = x
#         self.next = None

# class Solution(object):
#         """
#         :type n: int
#         :rtype: ListNode
#         """
class Solution(object):
#     # with O(n) space
#     index = []
#     while pos is not None:
#         index.append(pos)
#         pos = pos.next
#     ls = len(index)
#     if n == ls:
#         if ls > 1:
#             return index[1]
#         else:
#             return None
#     else:
#         index_pos = ls - n - 1
#         index[index_pos].next = index[index_pos + 1].next

# https://leetcode.com/discuss/86721/o-n-solution-in-java
return None
for i in range(n):
fast = fast.next
if fast is None:
while fast.next is not None:
fast = fast.next
slow = slow.next
curr = slow.next
slow.next = curr.next
``````

## List of all Remove Nth Node From End of List problems

Leetcode 019 Remove Nth Node From End of List problem solution in python3 with explanation. This is the best place to expand your knowledge and get prepared for your next interview.

After removing the second node from the end, the linked list becomes 1->2->3->5.

Note:

Given n will always be valid.

Could you do this in one pass?

class ListNode(object):

def __init__(self, x):

self.val = x

self.next = None

class Solution(object):

"""

:type n: int

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