A1074 Reversing Linked List (25分)未解决看懂

二、参考代码

#include<cstdio>
#include<iostream>
using namespace std;
int main(){
    int first, k, n, sum = 0;
    cin >> first >> n >> k;
    int temp, data[100010], next[100010], list[100010], result[100010];
    for(int i = 0; i < n; i++){
        cin >> temp;
        cin >> data[temp] >> next[temp];
    }
    while(first != -1){
        list[sum++] = first;
        first = next[first];
    }
    for(int i = 0; i < sum; i++) result[i] = list[i];
    for(int i = 0; i < (sum - sum%k); i++){
        result[i] = list[i / k * k + k - 1 - i % k];
    }
    for(int i = 0; i < sum - 1; i++){
        printf("%05d %d %05d\n", result[i], data[result[i]], result[i+1]);
    }
    printf("%05d %d -1", result[sum - 1], data[result[sum-1]]);
    return 0;
}

原文地址：https://www.cnblogs.com/tsruixi/p/12257482.html