1074. Reversing Linked List (25)

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本文介绍了一种链表处理算法，该算法可以按指定长度K对链表进行分组，并对每一组内的元素进行反转。通过使用顺序数组存储节点地址，结合标准库中的reverse函数实现了高效反转。

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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

将一个链表按照长度m倒转，注意每m个节点要倒转一下，而不仅仅是前m个要倒转。用val记录地址对应的值，用next记录下一个点的值。然后由给出的开始的节点开始遍历链表，用order数组储存节点地址的顺序。然后用reverse函数对order数组进行倒转操作，每m个倒转一下。最后按照倒转后的order输出结果即可。最后还要注意给出的节点不一定都在链表内，所以在遍历求得order数组的过程中要排除掉那些不在链表中的点，不然最后的case过不了。

代码：

#include <iostream>
#include <cstring>
#include <cstdlib>
#include <cstdio>
#include <vector>
#include <algorithm>
using namespace std;

const int SIZE=100005;

int main()
{
	int first,n,m;
	cin>>first>>n>>m;
	int val[SIZE];
	int next[SIZE];
	for(int i=0;i<n;i++)
	{
		int addr,v,naddr;
		cin>>addr>>v>>naddr;
		val[addr]=v;
		next[addr]=naddr;
	}
	int cur=first;
	vector<int>order;
	for(int i=0;i<n;i++)
	{
		if(cur==-1) break;
		order.push_back(cur);
		cur=next[cur];
	}
	n=order.size();
	int k=0;
	while(m!=0&&k+m<=n)
	{
		reverse(order.begin()+k,order.begin()+k+m);
		k+=m;
	}
	cout<<endl;
	for(int i=0;i<n;i++)
	{
		printf("%05d %d ",order[i],val[order[i]]);
		if(i==n-1)
		{
			printf("-1\n");
		}
		else
		{
			printf("%05d\n",order[i+1]);
		}
	}
}