Reversing Linked List

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 400000 4 9999900100 1 1230968237 6 -133218 3 0000099999 5 6823712309 2 33218

Sample Output:

00000 4 3321833218 3 1230912309 2 0010000100 1 9999999999 5 6823768237 6 -1

#include<iostream>#include<algorithm>#include <iomanip>using namespace std;int list[100010];int node[100010][2];//二维数组：行存放地址，列存放元素和下一个地址int main(){    int st,num,k;    cin>>st>>num>>k;    int address,data,next,i;    for(i=0;i<num;i++)    {        cin>>address>>data>>next;        node[address][0]=data;        node[address][1]=next;    }    int m=0,n=st;    while(n!=-1)//将地址存放在list中    {        list[m++]=n;        n=node[n][1];    }i=0;    while(i+k<=m)//反转list中的地址    {        reverse(list+i,list+i+k);        i=i+k;    }    for(i=0;i<m;i++)    {cout<<right<<setw(5)<<setfill('0')<<list[i]<<" ";//右对齐，元素长度为5，不足补0。这是因为00000的地址输出是0cout<<node[list[i]][0]<<" ";if(i!=m-1)cout<<right<<setw(5)<<setfill('0')<<list[i+1]<<'\n';elsecout<<-1<<'\n';//最后一个行最后一个输出    }}