PTA L2-022 C++ Realization of Rearranging Linked List

Given a singly linked list L1→L2→⋯→Ln−1→Ln, please write a program to rearrange the linked list as Ln→L1→Ln−1→L2→⋯. For example: Given that L is 1→2→3→4→5→6, the output should be 6→1→5→2→4→3.

Input format:

Each input contains 1 test case. The first line of each test case gives the address of the first node and the total number of nodes, that is, positive integer N (≤105). The address of the node is a 5-bit non-negative integer, and the NULL address is represented by −1.

Next there are N lines, each line format is:

Address Data Next

Among them Addressis the node address; Datait is the data saved by the node, which is a positive integer not exceeding 105; Nextit is the address of the next node. The topic guarantees that there are at least two nodes on the given linked list.

Output format:

For each test case, sequentially output the rearranged result linked list, each node on it occupies one line, and the format is the same as the input.

Input sample:

00100 6
00000 4 99999
00100 1 12309
68237 6 -1
33218 3 00000
99999 5 68237
12309 2 33218

Sample output:

68237 6 00100
00100 1 99999
99999 5 12309
12309 2 00000
00000 4 33218
33218 3 -1

Problem-solving ideas:

The simulation process is enough, but it should be noted that the node given in the title may not be in this linked list, so a counter cnt is needed to count the length of this linked list to re-correct the value of n.

Code example:

#include <bits/stdc++.h>

using namespace std;

struct stu
{
    int data; //数据
    int next; //后继节点
    int pre;  //前驱节点
}res[100020];

vector<int> ans; //存已经重排好的节点的地址

int main()
{
    int first_ip, n;
    cin >> first_ip >> n;
    
    
    for(int i = 0; i < n; i++)
    {
        int ip;
        cin >> ip;
        cin >> res[ip].data >> res[ip].next;
    }
    
    int cnt = 0; //计数器 数链表的长度
    int p = first_ip;
    int q = -1;
    
    //本循环的作用是形成前驱节点和计算链表的真正长度
    while(p != -1)
    {
        res[p].pre = q;
        q = p;
        p = res[p].next;
        cnt++;
    }
    
    n = cnt; //修正n的值
    p = first_ip;
    while(q != -1 && p != -1)
    {
        ans.emplace_back(q);
        ans.emplace_back(p);
        
        q = res[q].pre;
        p = res[p].next;
    }
    
    for(int i = 0; i < n - 1; i++)
    {
        printf("%05d %d %05d\n", ans[i], res[ans[i]].data, ans[i + 1]);
    }
    
    printf("%05d %d -1\n", ans[n - 1], res[ans[n - 1]].data);
    
    return 0;
}