原题地址：http://oj.leetcode.com/problems/remove-nth-node-from-end-of-list/

题意：

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.

这道题的含义是删除链表的倒数第n个节点。

解题思路：加一个头结点dummy，并使用双指针p1和p2。p1先向前移动n个节点，然后p1和p2同时移动，当p1.next==None时，此时p2.next指的就是需要删除的节点。

代码：

# 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
        p1=p2=dummy
        for i in range(n): p1=p1.next
        while p1.next:
            p1=p1.next; p2=p2.next
        p2.next=p2.next.next
        return dummy.next