数据结构-2019夏 02-线性结构3 Reversing Linked List (25 分)_02-线性结构3 reversing linked list (25 分) given a cons-优快云博客

本文链接：https://blog.youkuaiyun.com/Billccx/article/details/95003349

本文介绍了一种算法，用于实现单链表中每K个元素的逆序操作。通过实例演示了当K为3或4时，如何将链表1→2→3→4→5→6逆序为3→2→1→6→5→4或4→3→2→1→5→6。详细解析了输入规格，包括节点地址、数据和下一个节点的位置，以及输出规格，即按相同格式打印逆序后的链表。

摘要生成于 C知道，由 DeepSeek-R1 满血版支持，前往体验 >

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 (≤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

代码：

#include <stdio.h>
struct node{
	int data;
	int next;
};
struct str{
	int data;
	int address;
};
struct node link[100003];
struct str string[100003];
int main(){
	int n,k,start;
	scanf("%d %d %d",&start,&n,&k);
	int i;
	for(i=0;i<n;i++){
		int add,dat,nex;
		scanf("%d %d %d",&add,&dat,&nex);
		link[add].data=dat;
		link[add].next=nex;
	}
	int p=start;
	for(i=1;p!=-1;p=link[p].next,i++){
		string[i].data=link[p].data;
		string[i].address=p;
	}
	i--;
	n=i;
	int round=1;
	while(n>=k&&k!=1){
		n=n-k;
		for(i=0;i<k;i++){
			if(n==0&&i==k-1)
				printf("%05d %d -1\n",string[round*k-i].address,string[round*k-i].data);
			else if(i!=k-1){
				printf("%05d %d %05d\n",string[round*k-i].address,string[round*k-i].data,string[round*k-i-1].address);	
			}
			else{
				if(n>=k)
					printf("%05d %d %05d\n",string[round*k-i].address,string[round*k-i].data,string[round*k-i+k*2-1].address);
				else
					printf("%05d %d %05d\n",string[round*k-i].address,string[round*k-i].data,string[round*k-i+k].address);
			}
		}
		round++;
	}
	round--;
	for(i=round*k+1;n>0;n--,i++){
		if(n!=1)
			printf("%05d %d %05d\n",string[i].address,string[i].data,string[i+1].address);
		else
			printf("%05d %d -1\n",string[i].address,string[i].data);
	}	
}