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https://leetcode.com/problems/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.
题目：删除链表倒数第n个结点。
倒数如何实现?利用dummy结点和两个指针p1和p2,p1先行n步，然后p1和p2再一起走，当p1.next为空时即走完整个链表时，p2.next即为要删除的倒数第n个结点。
（设链表总共M个结点，需要找到倒数第n个，那么p2就需要正序走并经过M-n个结点，下一个就是所需的倒数第n个结点，这正序的M-n步怎么控制，就是通过p1这个指针）
1 # Definition for singly-linked list.
2 # class ListNode:
3 # def __init__(self, x):
4 # self.val = x
5 # self.next = None
6
7 class Solution:
8 # @param {ListNode} head
9 # @param {integer} n
10 # @return {ListNode}
11 def removeNthFromEnd(self, head, n):
12 dummy=ListNode(0)
13 dummy.next=head
14 p1=p2=dummy #p1,p2同时dummy出发
15 for i in range(n): #p1先走n步
16 p1=p1.next
17 while p1.next: #p1,p2同时走，直到p1走完链表
18 p1=p1.next
19 p2=p2.next
20 p2.next=p2.next.next #删去此时的p2.next
21 return dummy.next
Remove Nth Node From End of List
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原文地址：http://www.cnblogs.com/lzsjy421/p/4609565.html
