【PTA-A】1052 Linked List Sorting (25 分)（链表）

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本文介绍了一种基于链表结构的排序算法实现，通过遍历链表记录节点信息，然后对节点进行排序，最后重新构建排序后的链表。文章详细解释了算法步骤，包括节点的标记、排序过程以及输出格式。

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

思路：

1.创建结构体存储当前位置，key，下一个位置

2.从开头位置不断查找有效的下一位，直到碰到“-1”结束，记录该条链的长度，标记在链表中的节点

3.对节点先按是否有效排序，再按key值从小到大排序

4.输出，其中下一个位置指向下一个key的位置，最后输出-1

注意点：

1.利用%05d 输出五位数的地址，-1另外输出

2.输出的下一位置是下一节点的存储位置

3.如果没有有效节点，就输出“0 -1”

#include<iostream>
#include<string>
#include<cstring>
#include<algorithm>
using namespace std;
const int MAXN = 100010;
struct Node {
	int ad, key, next;
	int flag;
}no[MAXN];

bool cmp(Node a, Node b) {
	if (a.flag != b.flag)return a.flag > b.flag;
	else return a.key < b.key;
}
int main() {
	int n, start,temp;
	cin >> n >> start;
	for (int i = 0; i < n; i++) {
		cin >> temp;
		scanf("%d %d",&no[temp].key,&no[temp].next);
		no[temp].ad = temp;
	}
	int p = start,count=0;
	while (p != -1) {
		count++;
		no[p].flag = 1;
		p = no[p].next;
	}
	if (count == 0) {
		cout << "0 -1";
	}
	else {
		sort(no, no + MAXN, cmp);
		printf("%d %05d\n", count, no[0].ad);
		for (int i = 0; i < count-1; i++) {
			printf("%05d %d %05d\n", no[i].ad, no[i].key,no[i+1].ad);
		}
		printf("%05d %d -1", no[count - 1].ad, no[count - 1].key);
	}
	return 0;
}