# Leetcode: Remove Nth Node From End of List

Remove Nth Node From End of List Similar Problems:

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

For 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.
• Try to do this in one pass.

Github: code.dennyzhang.com

Credits To: leetcode.com

Leave me comments, if you have better ways to solve.

```## Blog link: https://code.dennyzhang.com/remove-nth-node-from-end-of-list
## Basic Ideas: Two pointers with distance of n+1
##              Since the head node might be deleted, we need a dummy node
##              One pass
##              1->2->3->4->5, n=2
##                    p  .  . q
##
##   Assumption: If n is bigger than the length of the list, do nothing
## Complexity: Time O(n), Space O(1)
# class ListNode(object):
#     def __init__(self, x):
#         self.val = x
#         self.next = None

class Solution(object):
"""
:type n: int
:rtype: ListNode
"""
dummyNode = ListNode(None)