02-线性结构3 Reversing Linked List（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 $K$ ($\le N$) 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

题目大意：

反转链表，每隔K个链表元素就反转K个链表，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.

思路：c++，algorithm中有很多有用的函数，例如reverse函数，反转函数.具体用法可以参考：https://blog.youkuaiyun.com/u012877472/article/details/49557077

反转容器的函数，用于数组需要在数组前加取地符，例如：reverse（&array[i],&array[j]）效果是将i到j的数组元素反转。

我们可以用一个数组将链表地址储存起来，按照顺序存起来。然后反转，得到一个反转之后地址的数组。这样在输出。

数据结构设计：用数组模拟，首地址可以用数组下标代替，node中只需要data和next，两个数据类型。

具体代码：

#include<iostream>
#include<algorithm>
#include<cstdio>
using namespace std;
typedef struct node
{
 int data;
 int next;
}node;
node array[100000];
int address[100000];
int main()
{
 int N;
 int First;
 int K;
 cin>>First>>N>>K;
 int d;
 for(int i=0;i<N;i++)
 {
  cin>>d;
  cin>>array[d].data>>array[d].next;
 }
 int p=First;
 int cnt=0;
 while(p!=-1)//注意可能有不在链上的多余结点 
 {
  address[cnt++]=p;
  p=array[p].next;
 }
 int i=0;
 while(i+K<=cnt)//同理 
 {
  reverse(&address[i],&address[i+K]);
  i+=K;
 }
 for(i=0;i<cnt-1;i++)//array[address[i]].next不可信，注意 
  printf("%05d %d %05d\n",address[i],array[address[i]].data,address[i+1]);
 printf("%05d %d %d",address[i],array[address[i]].data,-1);
 return 0;
}

代码看不懂，应该用手在纸上写写！