题目要求

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 a Next 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

解题思路

首先想到的是通过按照key对所有的node进行排序，然后在输出的时候确定其next地址。

编写完成之后，发现只通过了两组数据，还有三组数据没有通过。再结合之前做过的相关问题，想到不是所有的node都在该Linked List上。于是开辟另一个数组，用于存储在该Linked List上的所有的节点，再排序输出。

继续提交后发现，最后一组数据没有通过，而且时间只有1ms。遂想到，有可能是最后一组数据可能是Linked List长度为0的数据。于是对其进行特殊的处理，其head node的地址是-1.

代码

#include <cstdio>
#include <cstdlib>
#include <algorithm>

using namespace std;

typedef struct Node
{
	int addr, key, next;
}Node;

Node node[100000 + 10];
Node nodes[100000 + 10];

bool cmp(const Node &a, const Node &b)
{
	return a.key < b.key;
}

int main() 
{
	int n = 0, start;
	scanf("%d%d", &n, &start);
	int addr, key, next;
	for (int i = 0; i < n; i++)
	{
		scanf("%d%d%d", &addr, &key, &next);
		node[addr].addr = addr;
		node[addr].key = key;
		node[addr].next = next;
	}
	int cnt = 0;
	int cur = start;
	while (cur != -1)
	{
		nodes[cnt] = node[cur];
		cur = node[cur].next;
		cnt++;
	}
	n = cnt;
	sort(nodes, nodes + n, cmp);
	if (cnt == 0)
	{
		printf("%d -1\n", n);
		return 0;
	}
	printf("%d %05d\n", n, nodes[0].addr);
	for (int i = 0; i < n - 1; i++)
	{
		printf("%05d %d %05d\n", nodes[i].addr, nodes[i].key, nodes[i + 1].addr);
	}
	printf("%05d %d -1\n", nodes[n - 1].addr, nodes[n - 1].key);
	
	system("pause");
	return 0;
}