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

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

思路

• 題意: 對一個連結串列進行排序
• 解法:
  • 1️⃣用靜態連結串列存結點, 從頭指標開始遍歷並存入一個vector, 對vector進行排序, 再按照規定的格式輸出即可.
• ⚠️要區分實際連結串列長度和給你的結點數, 兩者不一定相等.
• ⚠️實際連結串列長度為0的時候, 輸出0 -1

#include<bits/stdc++.h>
using namespace std;
const int MAXN = 100010;
struct node
{
	int address;
	int data;
	int next;
	int id;
}LinkList[MAXN];
bool cmp(node a, node b)
{
	return a.data < b.data;
}
int main()
{
	int n, st;
	scanf("%d%d", &n, &st);
	int address, next;
	int data;
	for(int i=0;i<n;i++)
	{
		scanf("%d %d %d", &address, &data, &next);
		LinkList[address].address = address;
		LinkList[address].data = data;
		LinkList[address].next = next;
		LinkList[address].id = i;
	}	//靜態連結串列
	vector<node> v;
	int cur = st;
	while(cur != -1)
	{
		v.push_back(LinkList[cur]);
		cur = LinkList[cur].next;
	}
	sort(v.begin(), v.end(), cmp);
	if(v.size() == 0)
    {
        printf("0 -1\n");
        return 0;
    }
	printf("%d %05d\n", v.size(), v[0].address);
	for(int i=0;i<v.size();i++)
	{
		printf("%05d %d", v[i].address, v[i].data);
		if(i == v.size()-1)
			printf(" -1\n");
		else
			printf(" %05d\n", v[i+1].address);
	}
	return 0;
}

應用

https://pintia.cn/problem-sets/994805342720868352/problems/994805425780670464