【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.

 1 struct ListNode {

 2     int val;

 3     ListNode *next;

 4     ListNode(int x) : val(x), next(NULL) {}

 5     

 6 };

 7 ListNode *removeNthFromEnd(ListNode *head, int n)

 8 {

 9     ListNode *r = head, *s;

10     if (head == NULL)

11         return head;

12     int i = 0;

13     for (ListNode *p = head; p != NULL; p = p->next)

14     {

15         i++; 

16     }

17     if (i == n)  //删除第一个节点

18     {

19         ListNode *l = head->next;

20         free(head);

21         return l;

22     }

23     for (int num = 0; num < i - n - 1; num++)//找到要删节点的前一个节点

24     {

25         r = r->next;

26     }

27     ListNode *tmp = r->next;

28     r->next = r->next->next;

29     free(tmp);

30     return head;

31     

32 }