题目：

Given a linked list, remove the n th 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.

解题思路：

定义两个指针，让快、慢指针同时指向头结点；然后让快指针先走n步，在让两个指针同时走，直到快指针走到结尾。这时，慢指针指向的结点就是要删除的结点。

代码实现：

class Solution {
public:
    ListNode *removeNthFromEnd(ListNode *head, int n) {
        if(head == NULL)
            return NULL;
        if(n < 0)
            return head;
        //计算节点个数
        int count = 0;
        ListNode* cur = head;
        while(cur)
        {
            count++;
            cur = cur->next;
        }
        if(n > count)
            return head;
        ListNode* fast = head;
        ListNode* slow = head;
        while(n--)
        {
            fast = fast->next;
        }
        ListNode* pre = NULL;
        while(slow && fast)
        {
            pre = slow;
            slow = slow->next;
            fast = fast->next;
        }
        if(slow == head)
        {
            head = head->next;
        }
        else
            pre->next = slow->next;
        delete slow;
        return head;
    }
};