题目：

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.

给定一个单链表，移除倒数第n个Node，返回链表的head

思路：维护两个指针slow，fast，fast先移动n步，然后slow、fast同时移动，直到fast为空为止，删除slow后面的结点即可

代码：

1 /** 2  * Definition for singly-linked list. 3  * struct ListNode { 4  *     int val; 5  *     ListNode *next; 6  *     ListNode(int x) : val(x), next(NULL) {} 7  * }; 8  */ 9 class Solution {10 public:11     ListNode* removeNthFromEnd(ListNode* head, int n) {12         ListNode *preHead = new ListNode(0);13         ListNode *fast = preHead, *slow = preHead;14         slow->next = head;15         for(int i = 0; i <= n; ++i)16             fast = fast->next;17         while(fast)18         {19             fast = fast->next;20             slow = slow->next;21         }22         slow->next = slow->next->next;23         return preHead->next;24     }25 };