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

解答：这道题中，是否重复的判定可以用set来保存第一个数组中的值，如果遍历到的新数在set里面，那么便插入到第二个数组中。

AC代码如下：

#include<iostream>
#include<cstdio>
#include<vector>
#include<set>
using namespace std;

typedef struct{
	int key; 
	int nxt;
}node;

int main()
{
	vector<node> nodes(100005);
	int saddr, num;
	
	
	//输入数据 
	scanf("%d %d", &saddr, &num);
	for(int i = 0; i < num; ++i)
	{
		int addr, key, nxt;
		scanf("%d %d %d", &addr, &key, &nxt);
		nodes[addr].key = key;
		nodes[addr].nxt = nxt;
	}
	
	//生成resulting linked list 和 removed list
	vector<int> result, removed;
	set<int> keys;
	
	int ptr = saddr;
	while(ptr != -1)
	{
		//如果不重复 
		if(keys.find(abs(nodes[ptr].key)) == keys.end()){
			result.push_back(ptr);
			keys.insert(abs(nodes[ptr].key));
			//cout << "ptr: " << ptr << endl; 
		//如果重复 
		}else{  
			removed.push_back(ptr);
		}
		ptr = nodes[ptr].nxt;
	}
	result.push_back(-1);
	removed.push_back(-1);
	//输出结果
	for(int i = 0; i < result.size() - 1; ++i){
		if(i < result.size() - 2) printf("%05d %d %05d\n", result[i], nodes[ result[i] ].key, result[i+1]);
		else printf("%05d %d %d\n", result[i], nodes[ result[i] ].key, result[i+1]);
	}
	for(int i = 0; i < removed.size() - 1; ++i){
		if(i < removed.size() - 2) printf("%05d %d %05d\n", removed[i], nodes[ removed[i] ].key, removed[i+1]);
		else printf("%05d %d %d\n", removed[i], nodes[ removed[i] ].key, removed[i+1]);
	}
	return 0;
}