02-线性结构3 Reversing Linked List（25 分）

题目介绍

Given a constant K and a singly linked list L, you are supposed to reverse the links of every K elements on L. For example, given L being 1→2→3→4→5→6, if K=3, then you must output 3→2→1→6→5→4; if K=4, you must output 4→3→2→1→5→6.

Input Specification:

Each input file contains one test case. For each case, the first line contains the address of the first node, a positive N (≤10

​5

​​ ) which is the total number of nodes, and a positive K (≤N) which is the length of the sublist to be reversed. The address of a node is a 5-digit nonnegative integer, and NULL is represented by -1.

Then N lines follow, each describes a node in the format:

Address Data Next

where Address is the position of the node, Data is an integer, and Next is the position of the next node.

Output Specification:

For each case, output the resulting ordered linked list. Each node occupies a line, and is printed in the same format as in the input.

Sample Input:

00100 6 4

00000 4 99999

00100 1 12309

68237 6 -1

33218 3 00000

99999 5 68237

12309 2 33218

Sample Output:

00000 4 33218

33218 3 12309

12309 2 00100

00100 1 99999

99999 5 68237

68237 6 -1

本题的坑点在于，题目输入时是给了你数据的地址的，这时传统创建链表的方式就不能用，这时需要用数组进行存储数据，数组下标存储地址。

#include <stdio.h>

#define MAX_SIZE 100001

typedef struct Node* LNode;
struct Node
{
int addr;
int data;
int next_addr;
LNode next;
};
//反转单链表函数
LNode ListReverse(LNode head,int k);
//输出单链表
void PrintList(LNode head);

int main()
{
int first_addr,k=0;
int n =0;
int temp_addr;//输入的时候用
int i,j;
int num;//统计链表建好后，链表中节点的数目
int Data[MAX_SIZE];//存data值，节点位置作为索引
int Next_addr[MAX_SIZE];//存next_addr值，节点位置作为索引
scanf("%d %d %d",&first_addr,&n,&k);
struct Node a[n+1];
//    设置a[0]为头结点
a[0].next_addr = first_addr;
for(i = 0;i<n;i++)
{
scanf("%d",&temp_addr);
scanf("%d %d",&Data[temp_addr],&Next_addr[temp_addr]);
}
i =1;
//    将链表串起来
while(1)
{
if(a[i-1].next_addr == -1)
{
a[i-1].next = NULL;
num = i-1;
break;
}
a[i].addr = a[i-1].next_addr;
a[i].data = Data[a[i].addr];
a[i].next_addr = Next_addr[a[i].addr];
a[i-1].next = a+i;
i++;
}
//将反转的链表rp连接到数组a中
LNode p = a;
LNode rp = NULL;
if(k <= num)
{
for (i = 0;i<(num/k);i++)
{
rp = ListReverse(p,k);
p->next = rp;
p->next_addr = rp->addr;
for(j=0;j<k;j++)
{
//    移向下一段需要反转的子链表的头结点
p = p->next;
}
}
}
//打印链表
PrintList(a);
return 0;

}
LNode ListReverse(LNode head,int k)
{
int count = 1;
LNode new1,old,temp;
new1 = head->next;
old = new1->next;
while(count < k)
{
temp = old->next;
old->next = new1;
old->next_addr = new1->addr;
new1 = old;
old = temp;
count ++;
}
head->next->next = old;
if(old != NULL)
{
head->next->next_addr = old->addr;
}
else{
head->next->next_addr = -1;
}
return new1;
}
void PrintList(LNode head)
{
LNode p = head;
while(p->next != NULL)
{
p = p->next;
if(p->next_addr != -1)
printf("%.5d %d %.5d\n",p->addr,p->data,p->next_addr);
else
printf("%.5d %d %d\n",p->addr,p->data,p->next_addr);
}
}