PAT 1074. Reversing Linked List (25)

1074. Reversing Linked List (25)

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

以下为了简便就使用了一个vector，如果为了省内存可以省略

#include <iostream>
#include <cstdio>
#include <vector>

using namespace std;

struct Node
{
    int address;
    int value;
    int next;
};

Node list[100000];

int main()
{
    int start, listNum, length;
    cin >> start >> listNum >> length;

    for (int i = 0; i < listNum; i++)
    {
        Node tmp;
        cin >> tmp.address >> tmp.value >> tmp.next;
        list[tmp.address] = tmp;
    }
    vector<Node> vec;
    int i = start, cnt = 0;
    Node* sublist = new Node[length];
    while (i != -1)
    {
        sublist[cnt] = list[i];
        cnt++;
        if (cnt == length)
        {
            for (int j = cnt-1; j>=0; j--)
                vec.push_back(sublist[j]);
            cnt = 0;
        }    
        i = list[i].next;
    }
    for (int j = 0; j < cnt; j++)
        vec.push_back(sublist[j]);
    
    printf("%.05d %d ", vec[0].address, vec[0].value);
    for (int i = 1; i < vec.size(); i++)
        printf("%.05d\n%.05d %d ", vec[i].address, vec[i].address, vec[i].value);
    printf("-1");
}