题目：给出一个链表和一个数k，比如，链表为1→2→3→4→5→6，k=2，则翻转后2→1→6→5→4→3，若k=3，翻转后3→2→1→6→5→4，若k=4，翻转后4→3→2→1→6→5，用程序实现。
我的思路是：采用插入法翻转链表。
#include<stdio.h>
#include<stdlib.h>
typedef struct node
{
int data;
struct node *next;
};
void CreateList(node *head,int length);
void PrintList(node *head);
void RotateList(node *head, int from, int to);
void ShiftList(node *head, int k, int n);
int main()
{
int len,k;
node * head = (node*)malloc(sizeof(node));
printf("input the list length:\n");
scanf_s("%d", &len);
CreateList(head, len);
PrintList(head);
printf("input k:\n");
scanf_s("%d", &k);
ShiftList(head, k, len);
PrintList(head);
system("pause");
return 0;
}
void CreateList(node *head,int len)
{
node *p,*q;
head->data = NULL;
p = (node*)malloc(sizeof(node));
p = head;
for (int i = 0; i < len; i++)
{
q = (node*)malloc(sizeof(node));
scanf_s("%d", &(q->data));
q->next = NULL;
p->next = q;
p = p->next;
}
}
void PrintList(node *head)
{
node *temp = (node*)malloc(sizeof(node));
temp = head;
while (temp->next != NULL)
{
temp = temp->next;
printf("%d ", temp->data);
}
printf("\n");
}
void RotateList(node *head, int from, int to)
{
node *p, *q, *tail;
p = (node*)malloc(sizeof(node));
p = head;
for (int i = 1; i < from; i++)
{
p = p->next;
}
tail = (node*)malloc(sizeof(node));
tail = p;
p = p->next;
q = (node*)malloc(sizeof(node));
for (int i = from; i < to; i++)
{
q = p->next;
p->next = q->next;
q->next = tail->next;
tail->next = q;
}
}
void ShiftList(node*head, int k,int n)
{
RotateList(head, 1, k);
RotateList(head, k+1, n);
}