# https://leetcode.com/problems/remove-nth-node-from-end-of-list/

# 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.
#
# Follow up:
# Could you do this in one pass?

# Definition for singly-linked list.
# class ListNode(object):
#     def __init__(self, val=0, next=None):
#         self.val = val
#         self.next = next
class Solution(object):
    def removeNthFromEnd(self, head, n):
        """
        :type head: ListNode
        :type n: int
        :rtype: ListNode
        """
        if head.val is None or head.next is None:
            return None

        to_remove = head
        pointer = head
        for i in range(n):
            if pointer.next is None and i == n - 1:
                head = head.next
                return head
            pointer = pointer.next

        while pointer.next:
            pointer = pointer.next
            to_remove = to_remove.next

        # remove the node under to_remove
        # if to_remove.next is not None:
        to_remove.next = to_remove.next.next

        # to_remove = None

        return head