【LeetCode OJ】Remove Nth Node From End of List

题目：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.

 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 = ;
     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 = ; num < i - n - ; 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;

 }