Leetcode (Python): 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.

Solution:

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

class Solution:
    # @return a ListNode
    def removeNthFromEnd(self, head, n):
        dummy = ListNode(0);
        dummy.next = head;
        fast = dummy;
        for i in range(0,n):
            if fast != None:
                fast = fast.next
        if fast == None or n==0:
            return head
        
        slow = dummy
        while fast.next != None:
            fast = fast.next
            slow = slow.next
        slow.next = slow.next.next
        return dummy.next