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.

麻烦一点的解法，比较耗时间：

/**
 * 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 *p = head;
	int count = 0;
	while (p != NULL)
	{
		p = p->next;
		count++;
	}
	p = head;
	for (int i = 1; i < count - n; i++)
	{
		p = p->next;
	}

	if (n == count)
	{
		head = head->next;
		return head;
	}
	ListNode *q = p;
	q = p->next;
	p->next = q->next;
	return head;
    }
};

/**
 * 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 *slow, *quick, *pre;
	slow = quick = head;
	for (int i = 0; quick != NULL; i++)
	{
		pre = slow;
		if (i >= n)
		{
			slow = slow->next;
		}
		quick = quick->next;
	}
	if (slow == head)
	{
		head = head->next;
		return head;
	}
	pre->next = slow->next;
	free(slow);
	return head;
    }
};