原題連結：https://leetcode.com/problems/remove-nth-node-from-end-of-list/description/

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

class Solution:
    def removeNthFromEnd(self, head, n):
        """
        :type head: ListNode
        :type n: int
        :rtype: ListNode
        """
 

        previous = head
        current = head
        for i in range(n):
            current = current.next
        if not current:
            return head.next
        else:
            while current.next:
                current = current.next
                previous = previous.next
            previous.
 
next = previous.next.next
        return head