思路:
1、创建带头结点且头结点为空的单链表
2、通过“头插法”输入数据
3、循环获取链表长度。根据长度和倒数第几个节点K算出要输出的节点
#include <stdio.h>
struct ListNode
{
int m_nKey;
struct ListNode* m_pNext;
};
static struct ListNode* creatNode()
{
struct ListNode *head = (struct ListNode*)malloc(sizeof(struct ListNode));
if(!head)
{
return NULL;
}
head->m_pNext = NULL;
return head;
}
static int printf_node(struct ListNode *head)
{
struct ListNode *p = head->m_pNext;
int len_node = 0;
while(p != NULL)
{
printf("%d ", p->m_nKey);
p = p->m_pNext;
len_node++;
}
printf("\n");
//printf("len_node = [%d]\n", len_node);
return 0;
}
/* 尾插法 */
static int add_end(struct ListNode* head, int data)
{
struct ListNode* new = (struct ListNode*)malloc(sizeof(struct ListNode));
if(!new)
{
return -1;
}
new->m_nKey = data;
new->m_pNext = NULL;
struct ListNode *p = head;
while(p->m_pNext != NULL)
{
p = p->m_pNext;
}
p->m_pNext = new;
return 0;
}
/* 头插法 */
static int add_beg(struct ListNode* head, int data)
{
struct ListNode* new = (struct ListNode*)malloc(sizeof(struct ListNode));
if(!new)
{
return -1;
}
new->m_nKey = data;
new->m_pNext = head->m_pNext;
head->m_pNext = new;
return 0;
}
static int printf_index(struct ListNode *head, int end_index)
{
struct ListNode *p = head->m_pNext;
int len_node = 0;
while(p != NULL)
{
p = p->m_pNext;
len_node++;
}
//printf("len_node = [%d]\n", len_node);
p = head;
for(int i = 0; i < len_node-end_index+1; i++)
{
p = p->m_pNext;
}
printf("%d\n", p->m_nKey);
return 0;
}
int main()
{
int num = 0;
while(scanf("%d", &num) != EOF)
{
if(num >= 1 && num <= 1000)
{
struct ListNode* head = creatNode();
int data = 0;
for(int i = 0; i < num; i++)
{
scanf("%d", &data);
add_end(head, data);
}
//printf_node(head);
int k = 0;
scanf("%d", &k);
printf_index(head, k);
}
}
return 0;
}


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