PAT 1074 Reversing Linked List[链表][一般]

1074 Reversing Linked List (25)（25 分）

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 (<= 10^5^) 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

题目大意：给出一个链表，按照数k对其进行反转，并输出反转后的链表。

代码来自：https://www.liuchuo.net/archives/1910

#include <iostream>
#include<stdio.h>
using namespace std;
int main() {
    int first, k, n, sum = ;
    cin >> first >> n >> k;
    int temp, data[], next[], list[], result[];
    for (int i = ; i < n; i++) {
        cin >> temp;
        cin >> data[temp] >> next[temp];
    }
    while (first != -) {//还需要再次统计，并不是所有的节点都是有效的。
        list[sum++] = first;
        first = next[first];
    }
    for (int i = ; i < sum; i++) result[i] = list[i];//现在result里都已经是顺序排列了
    for (int i = ; i < (sum - sum % k); i++){
        result[i] = list[i / k * k + k -  - i % k];
        //printf("%d %d\n",i,i / k * k + k - 1 - i % k);
    }
    for (int i = ; i < sum - ; i++)
        printf("%05d %d %05d\n", result[i], data[result[i]], result[i + ]);
    printf("%05d %d -1", result[sum - ], data[result[sum - ]]);
    return ;
}

//并没有真正的使用结构体来构造链表。data也是使用地址下标来存储，十分简洁。

1.由于输入的节点可能存在异常的，所以又重新统计了一下。

2.但是那个result被反转赋值为list的下标公式可真难。

这个代码来自：

#include<iostream>
#include<algorithm>
using namespace std;
int list[];
int node[][];
int main()
{
    int st,num,r;
    cin>>st>>num>>r;
    int address,data,next,i;
    for(i=;i<num;i++)
    {
        cin>>address>>data>>next;
        node[address][]=data;
        node[address][]=next;
    }
    int m=,n=st;
    while(n!=-)
    {
        list[m++]=n;
        n=node[n][];
    }
    i=;
    while(i+r<=m)
    {
        reverse(list+i,list+i+r);
        i=i+r;
    }
    for (i = ; i < m-; i++)
    {
        printf("%05d %d %05d\n", list[i], node[list[i]][], list[i+]);
    }
    printf("%05d %d -1\n", list[i], node[list[i]][]);
}

//这个和上边的思想类似，都是使用数组来存储。

1.data是通过二维数组中的第一维来表示，（其实都一样），均是通过地址下标找到的。

2.使用了reverse函数，没想到可以用reverse函数，队医一维数组来说。

while(i+r<=m)
    {
        reverse(list+i,list+i+r);
        i=i+r;
    }

//厉害。