Reversing Linked List

02-线性结构3 Reversing Linked List（25 分）
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

java过不去
#include
#include
using namespace std;
  
struct data
{
int add;
int d;
int next;
};
  
int main()
{
int first,N,K,m=0,num=0;
cin>>first>>N>>K;
data a[100000];
data p[100000];
data aa[100000];
data res[100000];
  
int i,j;
for(i=0;i>a[i].add;
    cin>>a[i].d;
    cin>>a[i].next;
    p[a[i].add].d=a[i].d;
    p[a[i].add].next=a[i].next;
}
  
  
i=0;
while(first!=-1)
{
aa[i].add=first;
aa[i].d=p[first].d;
aa[i].next=p[first].next;
first=p[first].next;
num++;
i++;
}
  
N=num;//可能有节点不在链表上
/*赋值法for(i=0;i