PAT 1074. 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 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
#include<iostream>
#include<vector>
#define max 100000
using namespace std;
struct ele{
int data=0;
int next=0;
};
int start=0,N=0,interval=0,first=0,tag=0;
void reverse(vector<ele>& vi,int& s){
int inter=interval; int fir,sec,thi,temp,last=s;
if(tag!=0) s=vi[s].next;
temp=s; fir=s; sec=vi[fir].next; thi=vi[sec].next;
for(int i=1;i<=inter-1;i++){
vi[sec].next=fir;
fir=sec;
sec=thi;
thi=vi[thi].next;
}
vi[temp].next=sec;
if(tag!=0) vi[last].next=fir;
if(tag++==0)
first=fir;
}
int main(){
int cnt,address;
cin>>start>>N>>interval;
vector<ele> vi(max);
for(int i=0;i<N;i++){
scanf("%d",&address);
scanf("%d%d",&vi[address].data,&vi[address].next);
}
int begin=start;
while(begin!=-1){cnt++; begin=vi[begin].next;}
if(interval==1||cnt<interval) first=start;
else{
int s=start,times;
times=cnt/interval;
while(times--){
reverse(vi,s);
}
}
begin=first;
while(vi[begin].next!=-1){
printf("%05d %d %05d\n",begin,vi[begin].data,vi[begin].next);
begin=vi[begin].next;
}
printf("%05d %d %d\n",begin,vi[begin].data,vi[begin].next);
return 0;
}