Problem: 19. 删除链表的倒数第 N 个结点

思路

1. 知道链表的长度
2. 确定要删除的节点的位置
3. 处理特殊情况：
   1. 只有一个元素
   2. n == count

解题方法

由思路可知

Code

/*** Definition for singly-linked list.* struct ListNode {*     int val;*     struct ListNode *next;* };*/
struct ListNode* removeNthFromEnd(struct ListNode* head, int n) {struct ListNode *p = head, *q;int count = 0;while(p) {p = p->next;count++;}q = p = head;int postion = 0;while(postion < count-n-1 && p->next) {p = p->next;postion++;}if(postion == count-n-1) {q = p->next;p->next = p->next->next;} else if(n == count) {q = head;head = p->next;}free(q);return head;
}