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; 
}

posted @ 2018-02-01 15:55  A-Little-Nut  阅读(134)  评论(0编辑  收藏  举报