PAT甲级A1074 Reversing Linked List (25)

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最新推荐文章于 2024-10-31 22:45:34 发布

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本文介绍了一种链表操作方法——每K个元素进行反转，并提供了完整的C++实现代码。该方法适用于需要对链表进行周期性反转处理的问题。

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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 (<= 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

题意：给定一个链表，让你每k个节点反转一次，最后不足k个结点则不反转，然后按反转后的顺序输出结果。

思路：

1.将数据储存在数组中，然后从起始节点开始遍历，每k个结点反转一次，直接使用reverse函数进行反转操作（头文件是algorithm），最后输出List链表即可。

2.注意链表中存储的是地址，末位地址输出-1，注意特殊处理。

参考代码：

#include<cstdio>
#include<algorithm>
#include<vector>
using namespace std;
const int maxn=100010;
int main()
{
	int n,st,k,sum=0,addr,data[maxn],next[maxn];
	vector<int> List;
	scanf("%d%d%d",&st,&n,&k);
	for(int i=0;i<n;i++)	
	{
		scanf("%d",&addr);
		scanf("%d %d",&data[addr],&next[addr]);
	}
	while(st!=-1){
		List.push_back(st);
		sum++;
		st=next[st];
		if(k>1&&sum%k==0)
			reverse(List.begin()+sum-k,List.begin()+sum);
	}
	for(int i=0;i<sum-1;i++)
	{
		printf("%05d %d %05d\n",List[i],data[List[i]],List[i+1]);
	}
        printf("%05d %d -1\n", List[sum - 1], data[List[sum - 1]]);
	return 0;
}