PAT甲级1097 Deduplication on a Linked List (25 分)

最新推荐文章于 2020-09-04 17:24:55 发布

原创最新推荐文章于 2020-09-04 17:24:55 发布 · 171 阅读

0 ·

CC 4.0 BY-SA版权

文章标签：

#PAT

PAT甲级专栏收录该内容

158 篇文章

订阅专栏

本文提供了一道PAT在线评测系统中关于链表数据结构的题解，详细介绍了如何去除链表中绝对值重复的节点，并附上了完整的C++代码实现。通过使用静态数组和布尔数组来标记节点值的首次出现，实现了链表的去重并分别输出了去重后的链表和被移除的节点链表。

1097 Deduplication on a Linked List (25 分)

Given a singly linked list L with integer keys, you are supposed to remove the nodes with duplicated absolute values of the keys. That is, for each value K, only the first node of which the value or absolute value of its key equals K will be kept. At the mean time, all the removed nodes must be kept in a separate list. For example, given Lbeing 21→-15→-15→-7→15, you must output 21→-15→-7, and the removed list -15→15.

Input Specification:

Each input file contains one test case. For each case, the first line contains the address of the first node, and a positive N (≤105) which is the total number of nodes. The address of a node is a 5-digit nonnegative integer, and NULL is represented by −1.

Then N lines follow, each describes a node in the format:

Address Key Next

where Address is the position of the node, Key is an integer of which absolute value is no more than 104, and Next is the position of the next node.

Output Specification:

For each case, output the resulting linked list first, then the removed list. Each node occupies a line, and is printed in the same format as in the input.

Sample Input:

00100 5

99999 -7 87654

23854 -15 00000

87654 15 -1

00000 -15 99999

00100 21 23854

Sample Output:

00100 21 23854

23854 -15 99999

99999 -7 -1

00000 -15 87654

87654 15 -1

题目大意：

给出一个起始地址以及n结点，随后n行即是结点地址、值以及下个结点的地址。要求将绝对值相同的结点剔除出去，只保留第一个结点。要求输出新链表以及删除的链表。

思路：

静态数组问题：用数组下标作为结点地址，存储值与下个结点的地址。

其次遍历数组，用一个数组作为该结点值是否初次出现，存入输出链表ans。否则存入删除链表dl。

有个特别重要的，而且容易的忽略的知识点，关于PAT段错误的深层原因：https://blog.youkuaiyun.com/qq_42306885/article/details/99606600

参考代码：

#include<cstdio>
#include<vector>
#include<algorithm>
using namespace std;
struct node{
	int v, ad;
}temp[1000000];
vector<node>ans, dl;
bool ft[100000];
int main(){
	int st, n;
	scanf("%d%d", &st, &n);
	for(int i = 0; i < n; ++i){
		int ad;
		scanf("%d", &ad);
		scanf("%d%d", &temp[ad].v, &temp[ad].ad);
	}
	while(st != -1){
		if(!ft[abs(temp[st].v)]){
			ft[abs(temp[st].v)] = true;
			ans.push_back({temp[st].v, st});
		}else	dl.push_back({temp[st].v, st});
		st = temp[st].ad;
	}
	for(int i = 0; i + 1< ans.size(); ++i)	printf("%05d %d %05d\n", ans[i].ad, ans[i].v, ans[i + 1].ad);
	if(ans.size() != 0)		printf("%05d %d -1\n", ans[ans.size() - 1].ad, ans[ans.size() - 1].v);
	for(int i = 0; i + 1< dl.size(); ++i)	printf("%05d %d %05d\n", dl[i].ad, dl[i].v, dl[i + 1].ad);
	if(dl.size() != 0)		printf("%05d %d -1", dl[dl.size() - 1].ad, dl[dl.size() - 1].v);
	return 0;
}