给定一个链表，删除链表中倒数第n个节点，返回链表的头节点。

样例

给出链表1->2->3->4->5->null和 n = 2.
删除倒数第二个节点之后，这个链表将变成1->2->3->5->null.
解题思路：统计出链表的长度，然后就很简单了
/**
* Definition of ListNode
* class ListNode {
* public:
* int val;
* ListNode *next;
* ListNode(int val) {
* this->val = val;
* this->next = NULL;
* }
* }
*/
class Solution {
public:
/**
* @param head: The first node of linked list.
* @param n: An integer.
* @return: The head of linked list.
*/
ListNode *removeNthFromEnd(ListNode *head, int n) {
// write your code here
ListNode *p = head;
ListNode *p1 = head;
int num = 0;
while(p != NULL)
{
num += 1;
p = p->next;
}
if (num == n)
{
return head->next;
}
num = num - n - 1;
int numTemp = 0;
while(num != numTemp)
{
numTemp += 1;
head = head->next;
}
head->next = head->next->next;
return p1;
}
};