PAT Advanced 1052 Linked List Sorting (25) [链表]

题目

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

题目分析

已知N个节点，将节点按照值排序（升）

解题思路

1. 定义结构体数组，保存所有结点信息，节点中定义标记属性flag（初始化为false）
2. 遍历所有结点信息，将出现的节点flag标记为true，并定义计数器统计链表中结点数count
3. 遍历链表结点，修改next值为下一个结点

易错点

1. 题目中不容易分析出结点中有无效结点
2. 地址格式必须为"%05d"
3. 若结点数为0，打印"0 -1"

知识点

1. 将无效结点排序到最后的办法
   
   !a.flag || !b.flag ? a.flag > b.flag//降序true=1在前，false=0在后

Code

Code 01

#include <iostream>
#include <algorithm>
using namespace std;
struct NODE {
	int address, key, next;
	bool flag;
} node[100000];
int cmp1(NODE a, NODE b) {
	return !a.flag || !b.flag ? a.flag > b.flag : a.key < b.key;
}
int main() {
	int n, cnt = 0, s, a, b, c;
	scanf("%d%d", &n, &s);
	for(int i = 0; i < n; i++) {
		scanf("%d%d%d", &a, &b, &c);
		node[a] = {a, b, c, false};
	}
	for(int i = s; i != -1; i = node[i].next) {
		node[i].flag = true;
		cnt++;
	}
	if(cnt == 0) {
		printf("0 -1");
	} else {
		sort(node, node + 100000, cmp1);
		printf("%d %05d\n", cnt, node[0].address);
		for(int i = 0; i < cnt; i++) {
			printf("%05d %d ", node[i].address, node[i].key);
			if(i != cnt - 1)
				printf("%05d\n", node[i + 1].address);
			else
				printf("-1\n");
		}
	}
	return 0;
}