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 (≤105) 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
AC代码：
//Reversing Linked List (25)
#include<iostream>
#include<stdio.h>
#include<algorithm>
using namespace std;
#define MAXSIZE 1000010
struct List //存储数据和链表下一项的地址
{
int Data;
int Next;
}List[MAXSIZE];
int result[MAXSIZE];
int main()
{
int first,n,k;
int address,data,next;
cin>>first>>n>>k;
for(int i=0;i<n;i++)//按照地址存储信息
{
cin>>address>>data>>next;
List[address].Data=data;
List[address].Next=next;
}
int j=0,p;
p=first;//指向首节点
while(p!=-1)//将链接好的节点存储在一个顺序数组中
{
result[j++]=p;
p=List[p].Next;
}
int i=0;
while(i+k<=j)//翻转i~i+k之间的所有元素，即实现按k翻转
{
reverse(&result[i],&result[i+k]);
i=i+k;
}
for(i=0;i<j-1;i++)//result数组中不存储数据，只存储地址，按照result所给地址输出List的内容
printf("%05d %d %05d\n",result[i],List[result[i]].Data,result[i+1]);
printf("%05d %d -1\n",result[i],List[result[i]].Data);
return 0;
}
//string str="C++REVERSE";
//reverse(str.begin(),str.end());//str结果为ESREVER++C
//cout<<str;