【LeetCode OJ】Remove Nth Node From End of List

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.

struct ListNode {
int val;
ListNode *next;
ListNode(int x) : val(x), next(NULL) {}

};
ListNode *removeNthFromEnd(ListNode *head, int n)
{
ListNode *r = head, *s;
if (head == NULL)
return head;
int i = 0;
for (ListNode *p = head; p != NULL; p = p->next)
{
i++;
}
if (i == n)  //删除第一个节点
{
ListNode *l = head->next;
free(head);
return l;
}
for (int num = 0; num < i - n - 1; num++)//找到要删节点的前一个节点
{
r = r->next;
}
ListNode *tmp = r->next;
r->next = r->next->next;
free(tmp);
return head;

}