LeetCode:Remove Nth Node From End of List

问题描述：

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.
思路：

首先用一个循环将一个指针指向第n个节点；然后再用一个循环将一个指针指向被删除节点的前一个节点，另一个指针指向尾节点，然后判断从当前节点是否是头节点，如果是，则将头节点指向当前节点的下一个节点，如果不是，则将指定位置节点删除。

代码：

/**
* 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) {
if (head == NULL)
return NULL;

ListNode *pre = NULL;
ListNode *p = head;
ListNode *q = head;
for(int i = 0; i < n - 1; i++)
q = q->next;

while(q->next)
{
pre = p;
p = p->next;
q = q->next;
}

if (pre == NULL)
{
head = p->next;
delete p;
}
else
{
pre->next = p->next;
delete p;
}
return head;
}
};