✍个人博客:https://blog.csdn.net/Newin2020?spm=1011.2415.3001.5343
专栏地址:PAT题解集合
原题地址:题目详情 - 1052 Linked List Sorting (pintia.cn)
中文翻译:链表排序
专栏定位:为想考甲级PAT的小伙伴整理常考算法题解,祝大家都能取得满分!
❤️如果有收获的话,欢迎点赞收藏,您的支持就是我创作的最大动力
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], andNext
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
第一行首先包含一个整数 N N N,表示总节点数量,然后包含链表头节点的地址。
接下来 N N N 行,每行描述一个节点的信息,格式如下:
Address Key Next
其中 Address
是节点地址,Key
值是一个整数,Next
是下一个节点的地址。
地址都是 5
位正整数,可能有前导 0
。
保证各节点的 Key
值互不相同,且链表中不存在环形结构。
第一行首先输出一个整数表示链表的总节点数量,然后输出排序后的链表头节点地址。
接下来若干行,从头节点开始,依次输出每个节点的信息,格式与输入相同。
注意:
−1
)。key
是结点的地址,value
是结点的三个信息,用结构体存储。next
要输出 -1
。#include
using namespace std;
const int N = 100010;
struct Node {
string address;
int key;
string next;
//设定排序规则
bool operator <(const Node& x)const
{
return key < x.key;
}
};
int main()
{
int n;
string head;
cin >> n >> head;
//输入结点信息
unordered_map<string, Node> num;
for (int i = 0; i < n; i++)
{
char address[10], next[10];
int key;
scanf("%s %d %s", address, &key, next);
num[address] = { address,key,next };
}
//将存在链表中的筛选出来
vector<Node> res;
for (string i = head; i != "-1"; i = num[i].next) res.push_back(num[i]);
cout << res.size();
if (res.empty()) puts(" -1");
else
{
sort(res.begin(), res.end()); //对链表中结点的key值从小到大进行排序
cout << " " << res[0].address << endl;
for (int i = 0; i < res.size(); i++)
{
if (i + 1 == res.size()) //如果没有遍历到最后一个结点
printf("%s %d -1\n", res[i].address.c_str(), res[i].key);
else //如果遍历到最后一个结点
printf("%s %d %s\n", res[i].address.c_str(), res[i].key, res[i + 1].address.c_str());
}
}
return 0;
}