本文为senlie原创，转载请保留此地址：http://blog.csdn.net/zhengsenlie

   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.

题意：移除链表的倒数第n个元素
思路：
两个指针p, q，
p先走n步，然后p，q一起走，当p走到尾的时候，q->next就是要删除的节点
复杂度： 时间O(n)，空间O(1)

/**
 * Definition for singly-linked list.
 * struct ListNode {
 *     int val;
 *     ListNode *next;
 *     ListNode(int x) : val(x), next(NULL) {}
 * };
 */
class Solution {
public:
    ListNode *removeNthFromEnd(ListNode *head, int n) {
    	ListNode dummy(0);
    	dummy.next = head;
    	ListNode *p, *q;
    	p = q = &dummy;
    	while(n--) p= p->next;
    	while(p->next){
    		p = p->next;
    		q = q->next;
    	}
    	ListNode *to_delete = q->next;
    	q->next = to_delete->next;
    	delete to_delete;
    	return dummy.next;
    }
};