1052 Linked List Sorting （25 分）链表排序

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 $N(<10^5)$ 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

　　题目大意： 给你N个链表的结点，以及开始的地址，然后对链表进行排序，并按格式输出排序后的链表。
　　解题思路： 这道题不是很难，乙级有一道和这类似的，但是这道题有两个坑，一个是给你的元素，一部分不是链表上的，所以要处理一下；一个是可能存在空链的情况，注意到这种情况就好了。
　　还有一个投机取巧的方式，排序完之后，后一个结点的地址就是前一个结点的next，直接输出即可，不必重新建链。

/*
** @Brief:No.1052 of PAT advanced level.
** @Author:Jason.Lee
** @Date:2018-12-21
** @Solution: https://blog.csdn.net/CV_Jason/article/details/85169580
*/
#include
#include
#include
using namespace std;

struct node{
	int addr;
	int data;
	int next;
};

node nnn[100010];

int main(){
	int N,index,start;
	while(cin>>N){
		cin>>start;
		vector<node> list;
		node input{-1,0,-1};
		fill(nnn,nnn+100010,input);
		for(int i=0;i<N;i++){
			cin>>input.addr>>input.data>>input.next;
			nnn[input.addr] = input;
		}
		index = start;
		while(index!=-1){
			if(nnn[index].addr == -1){
				break;
			}
			list.push_back(nnn[index]);
			index = nnn[index].next;
		}
		if(list.size()!=0){
			sort(list.begin(),list.end(),[](node a,node b){return a.data<b.data;});
			printf("%d %05d\n",list.size(),list[0].addr);
			printf("%05d %d ",list[0].addr,list[0].data);
			for(int i=1;i<list.size();i++){
				printf("%05d\n%05d %d ",list[i].addr,list[i].addr,list[i].data);
			}
			printf("-1\n");
		}else{
			printf("0 -1\n");
		}
	}
	return 0;
}

总结

　　最后一个测试点是空链，最开始没想到，折腾了半天……所以这就陷入了一个很纠结的问题，2分的一个测试点，花了一个多小时，……唉……