题目要求：

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.

输入格式：

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. 

输出格式： 

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. 

 样例输入：

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

样例输出： 

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

题目翻译：

题解： 

        思路如注释所示，可通过所有测试点。

#include<bits/stdc++.h>

using namespace std;

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

int main(){
	int start,N,K;
	cin>>start>>N>>K;
	
	const int MAXN = 100000;
	Node nodes[MAXN]; //存储所有节点
	int result[MAXN]; //存储逆序后的链表地址
	
	for(int i=0;i<N;i++){
		int address,data,next;
		cin>>address>>data>>next;
		nodes[address] = {address,data,next};
	} 
	
	//重新构造链表 
	int count = 0;
	int current = start;
	while(current != -1){
		result[count++] = current;
		current = nodes[current].next;		
	}
	
	//进行链表逆转
	//直接交换链表的地址值进行逆置 
	for(int i = 0;i+K <= count;i += K){
		for(int j = 0;j<K/2;j++){
			int tmp = result[i+j];
			result[i+j] = result[i+K-1-j];
			result[i+K-1-j] = tmp;   	
		} 
	}
	
	//输出结果
	for(int i=0;i<count;i++){
		int address = result[i];
		int data = nodes[address].data;
		int next = (i!=count-1) ? result[i+1]:-1;
		
		printf("%05d %d ",address,data);
		if(next == -1)
			printf("-1\n");
		else
			printf("%05d\n",next);
	}
	 
	return 0;
} 

关于此题注意要点：

        1. 对链表的存储，使用了结构数组的方法，这样可以直接存储并操作其地址，这样我们在进行逆转时直接交换两个节点的地址即可。

        2.使用result数组对链表中的元素进行一个有序的存储，这样可以在交换个别节点地址时保证链表本身的有序性。

闲谈：

        因为我本身不是专业参加竞赛选手，自己编码能力暂时有限，我本身想用陈姥姥上课方法实现此题。理解了她讲的大概思路，但是总是细节处理不当，也没有找到很清晰的题解，故而换了一种思路求解，可能实现过程更加简单一点，也能解出来题，但是对编码能力锻炼不强，点到即止吧。（上周发烧停更了一周，有些懈怠，痛定思痛继续打卡）

        咳咳项目突然催进度了，等我把数据库学完再回归:(