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.

use a stack to record nodes. and pop out n+1 times to get the n+1th node from end. Then, delete the nth node

# 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(0)
        dummy.next = head
        stack = []
        l = -1
        current = dummy
        while current:
            stack.append(current)
            current = current.next
            l += 1
        node = 0
        for _ in xrange(n+1):
            node = stack.pop()
        node.next = node.next.next
        return dummy.next