Remove Nth node from End of List

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.

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

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

class Solution(object):
    def removeNthFromEnd(self, head, n):
        """
        :type head: ListNode
        :type n: int
        :rtype: ListNode
        """
        if head == None:
            return head
        else:
            dummy = ListNode(0)
            dummy.next = head
            fast = dummy
            slow = dummy
            for i in range(n):
                fast = fast.next

            while fast.next != None:
                fast = fast.next
                slow = slow.next

            slow.next = slow.next.next

            return dummy.next