1052. Linked List Sorting (25)

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最新推荐文章于 2025-08-22 13:08:41 发布

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分类专栏： pat-advanced 文章标签： sort list

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本文介绍了一种基于结构体的链表排序方法，通过遍历链表收集节点信息，并使用sort函数对节点按key值升序排列。文章提供了完整的C++实现代码。

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A linked list consists of a series of structures, which are not necessarily adjacent in memory. We assume that each structure contains an integer key and aNext pointer to the next structure. Now given a linked list, you are supposed to sort the structures according to their key values in increasing order.

Input Specification:

Each input file contains one test case. For each case, the first line contains a positive N (< 10⁵) and an address of the head node, where N is the total number of nodes in memory and the address of a node is a 5-digit positive integer. NULL is represented by -1.

Then N lines follow, each describes a node in the format:

Address Key Next

where Address is the address of the node in memory, Key is an integer in [-10⁵, 10⁵], and Next is the address of the next node. It is guaranteed that all the keys are distinct and there is no cycle in the linked list starting from the head node.

Output Specification:

For each test case, the output format is the same as that of the input, where N is the total number of nodes in the list and all the nodes must be sorted order.

Sample Input:

5 00001
11111 100 -1
00001 0 22222
33333 100000 11111
12345 -1 33333
22222 1000 12345

Sample Output:

5 12345
12345 -1 00001
00001 0 11111
11111 100 22222
22222 1000 33333
33333 100000 -1

要求给一个链表排序，其中数值都是唯一的。这里用sort函数对结构体就能完成排序的问题，然后依次输出当前点地址，当前点的数值和当前点的下一个点的地址。不过要注意的是给出的点不一定全部在这个链表中，要先从头节点开始先遍历一遍链表，把节点信息保存好再进行排序。还要注意的是链表是空的情况。

代码：

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

struct node
{
	int addr;
	int key;
	node(int a,int k):addr(a),key(k){}
};

bool cmp(const node &n1,const node &n2)
{
	return n1.key<n2.key;
}

int main()
{
	int n,first;
	scanf("%d %d",&n,&first);
	int next[100001];
	int keys[100001];
	memset(next,-1,100001);
	vector<node>list;
	for(int i=0;i<n;i++)
	{
		int from,key,to;
		scanf("%d %d %d",&from,&key,&to);
		if(from>=0&&from<100000)
		{
			next[from]=to;
			keys[from]=key;
		}
	}
	int cur=first;
	while(cur>=0)
	{
		list.push_back(node(cur,keys[cur]));
		cur=next[cur];
	}
	
	if(list.empty())
	{
		if(first==-1) 
			printf("0 -1\n");
		else
			printf("0 %5d\n",first);
		return 0;
	}
	sort(list.begin(),list.end(),cmp);
	printf("%d %05d\n",list.size(),list[0].addr);
	for(int i=0;i<list.size();i++)
	{
		if(i<list.size()-1) printf("%05d %d %05d\n",list[i].addr,list[i].key,list[i+1].addr);
		else printf("%05d %d -1\n",list[i].addr,list[i].key);
	}
}