标签：
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.
AC代码：
#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): front = behind = head for _ in xrange(n): front = front.next if front == None: return behind.next while front.next: front = front.next behind = behind.next behind.next = behind.next.next return head
思路：
因为是移除从后面数第n个，所以用两个指针，第一个先移动n次，然后两个指针一起移动，快的指针移动到最后的时候，慢的指针正好移动到离队尾差n个。
19. Remove Nth Node From End of List
标签：
原文地址：http://www.cnblogs.com/zhuifengjingling/p/5222139.html
