PAT-1097 Deduplication on a Linked List

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本文介绍了一个算法问题：如何从单链表中去除绝对值重复的节点，并将这些节点分离成一个新的链表。通过使用哈希表来跟踪已出现的值，并借助向量来组织输出，实现了原链表和重复元素链表的有效构建。

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1097 Deduplication on a Linked List (25)（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 L being 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 (<= 10^5^) 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 10^4^, 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

可以钻空子的题，将链表中绝对值重复的项拎出来（保留第一个）连成新的链表，输出原链表和重复元素链表。为了方tou便lan，用hashmap模拟内存寻址，pair<>存储数据，set判断是否重复，vector顺序装入要输出的元素，分别输出就行了。

#include<stdio.h>
#include<unordered_map>
#include<utility>
#include<vector>
#include<set>
using namespace std;

unordered_map<int,pair<int,int> > mem;
int main(){
	int head,n;
	scanf("%d %d",&head,&n);
	int addr;//,v,next;
	for(int i=0;i<n;i++){
		pair<int,int> p;
		scanf("%d %d %d",&addr,&p.first,&p.second);
		mem[addr]=p;
	}
	set<int> s;vector<pair<int,pair<int,int> > > v1,v2;
	for(addr=head;addr!=-1;addr=mem[addr].second){
		if(s.find(mem[addr].first)==s.end()){//show in first time
			s.insert(mem[addr].first);
			s.insert(-mem[addr].first);
			v1.push_back(pair<int,pair<int,int> >(addr,mem[addr]));
		}else{
			v2.push_back(pair<int,pair<int,int> >(addr,mem[addr]));
		}
	}
	auto it=v1.begin();
	if(v1.size()>0){
		for(;it!=v1.end()-1;it++){
			printf("%05d %d %05d\n",it->first,it->second.first,(it+1)->first);
		}
		printf("%05d %d %d\n",it->first,it->second.first,-1);
	}
	it=v2.begin();
	if(v2.size()>0){
		for(;it!=v2.end()-1;it++){
			printf("%05d %d %05d\n",it->first,it->second.first,(it+1)->first);
		}
		printf("%05d %d %d\n",it->first,it->second.first,-1);
	}
	return 0;
}