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


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.


Given n will always be valid.

Follow up:

Could you do this in one pass?


# Definition for singly-linked list.
# class ListNode:
#     def __init__(self, x):
#         self.val = x
# = None

class Solution:
    def removeNthFromEnd(self, head, n):
        :type head: ListNode
        :type n: int
        :rtype: ListNode
        nth_prev = head
        tail = head
        for i in range(0, n):
            tail =

        if tail is None:
            head =
            return head

        while is not None:
            nth_prev =
            tail = =
        return head