#1074. Reversing Linked List【链表】

原题链接

Problem Description：

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 $N$ ($\le 10^5$) which is the total number of nodes, and a positive $K$ ($\le N$) 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

Problem Analysis:

这里有一个很巧妙的方法，我们可以不直接对链表进行反转，转而将原有链表中的每一个节点的地址都存入一个 vector<int> 内，然后按照间隔 K 将 vector 反转，然后依次更新每个点的 next 指向下一个节点的地址，然后进行输出即可。

Code

#include <iostream>
#include <algorithm>
#include <cstring>
#include <vector>

using namespace std;

const int N = 1e5 + 10;

int n, m;
int h, e[N], ne[N];

int main()
{
    
       
    scanf("%d%d%d", &h, &n, &m);

    for (int i = 0; i < n; i ++ )
    {
    
    
        int address, data, next;
        scanf("%d%d%d", &address, &data, &next);
        e[address] = data, ne[address] = next;
    }

    vector<int> q;
    for (int i = h; ~i; i = ne[i]) q.push_back(i);

    for (int i = 0; i + m - 1 < q.size(); i += m)
        reverse(q.begin() + i, q.begin() + i + m);

    for (int i = 0; i < q.size(); i ++ )
    {
    
    
        printf("%05d %d ", q[i], e[q[i]]);
        if (i + 1 == q.size()) puts("-1");
        else printf("%05d\n", q[i + 1]);
    }

    return 0;
}