Reversing Linked List

最新推荐文章于 2022-03-21 23:03:24 发布

转载最新推荐文章于 2022-03-21 23:03:24 发布 · 394 阅读

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#c语言

本文介绍了一个算法问题：给定一个单链表L和一个常数K，每K个元素反转一次链表。输入包含链表的起始地址、节点总数及K的值，输出反转后的链表。文章提供了一个C语言实现的例子。

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Reversing Linked List

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 (<= 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.

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

#include<iostream>
#include<algorithm>
using namespace std;
int node[100005][2];
int list[100005];
int main()
{
	int start,num,length;
	int i;
	int add,data,next;
	while(cin>>start>>num>>length)
	{
		for(i=0;i<num;i++)
		{
			cin>>add>>data>>next;
			node[add][0]=data;
			node[add][1]=next;
		}
		int m=0,n=start;
		while(n!=-1)
		{
			list[m++]=n;
			n=node[n][1];
		}
		i=0;
		while(i+length<=m)
		{
			reverse(list+i,list+i+length);
			i=i+length;
		}
		for(i=0;i<m-1;i++)
			printf("%05d %d %05d\n",list[i],node[list[i]][0],list[i+1]);
		printf("%05d %d -1\n",list[i],node[list[i]][0]);
	}
	return 0;
}

关于C语言的，%5d是什么意思？？？
输出带符号的整数，需要占用5个字符位置, 不够则空格补齐. %5d是右对齐，%-5d是左对齐.
%05d是不足五位的前面补0