19. Remove N-th node from the end of linked 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, x):
#         self.val = x
#         self.next = None

class Solution(object):
    def removeNthFromEnd(self, head, n):
        """
        :type head: ListNode
        :type n: int
        :rtype: ListNode
        """
        dummy = ListNode(-217)
        dummy.next = head
        fast = dummy
        slow = dummy

        for i in range(n + 1):
        	fast = fast.next

        while(fast):
        	fast = fast.next
        	slow = slow.next

        slow.next = slow.next.next
        return dummy.next

Several things to consider:

1. We actually need to find the (n + 1)-th node from the end for deleting the n-th node.

2. How many steps fast should move ahead? n + 1 for either having dummy node or not.

3. Dummy node is necessary for removing the 1st node (the length-th node from end).