1097 Deduplication on a Linked List-PAT甲级_1097 deduplication柳婼-优快云博客

本文链接：https://blog.youkuaiyun.com/amf12345/article/details/98176597

本文介绍了一种处理单链表的方法，通过去除绝对值重复的键值，并将这些键值分离到另一个链表中。文章详细解释了如何使用data和nex数组处理节点值和地址，以及如何利用unordered_set和vector数组进行去重和分离操作。

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题目描述

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 L being 21→-15→-15→-7→15, you must output 21→-15→-7, and the removed list -15→15.

输入描述:

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.

输出描述:

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.

输入例子:

00100 5
99999 -7 87654
23854 -15 00000
87654 15 -1
00000 -15 99999
00100 21 23854

输出例子:

00100 21 23854
23854 -15 99999
99999 -7 -1
00000 -15 87654
87654 15 -1

链表的处理。选用data数组与nex数组用来处理当前节点的值与下一个节点的地址，用两个vector数组分别来存入没有重复键值以及是重复键值的，注意输出的处理。

满分代码如下：

#include<bits/stdc++.h>
using namespace std;
const int N=100100;
int data[N],nex[N];
unordered_set<int>st;
vector<int>aa;//非重复的键值 
vector<int>bb;
int main(){
	ios::sync_with_stdio(false);
	cin.tie(0);
	cout.tie(0);
	int n,first;
	cin>>first>>n;
	int address;
	for(int i=1;i<=n;i++){
		cin>>address;
		cin>>data[address]>>nex[address];
	}
	for(;first!=-1;first=nex[first]){
		if(st.find(abs(data[first]))==st.end()){
			st.insert(abs(data[first]));
			aa.push_back(first);
		}else{
			bb.push_back(first);
		}
	}
	for(int i=0;i<aa.size();i++){
		if(i<aa.size()-1){
			printf("%05d %d %05d\n",aa[i],data[aa[i]],aa[i+1]);
		}else{
			printf("%05d %d -1\n",aa[i],data[aa[i]]);
		}
	}
	for(int i=0;i<bb.size();i++){
		if(i<bb.size()-1){
			printf("%05d %d %05d\n",bb[i],data[bb[i]],bb[i+1]);
		}else{
			printf("%05d %d -1\n",bb[i],data[bb[i]]);
		}
	}
	return 0;
}