Day 52 B. Sorted Adjacent Differences

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本文介绍了一种算法，用于解决给定整数数组，通过重新排列元素来确保相邻元素差值按绝对值递减。题目要求找到一种可能的排列方式，即使数组中存在重复数字。博主提供了C++代码实现并给出了两个示例。

Problem
You have array of n numbers a1,a2,…,an.

Rearrange these numbers to satisfy |a1−a2|≤|a2−a3|≤…≤|an−1−an|, where |x| denotes absolute value of x. It’s always possible to find such rearrangement.

Note that all numbers in a are not necessarily different. In other words, some numbers of a may be same.

You have to answer independent t test cases.

Input
The first line contains a single integer t (1≤t≤10^4) — the number of test cases.

The first line of each test case contains single integer n (3≤n≤10^5) — the length of array a. It is guaranteed that the sum of values of n over all test cases in the input does not exceed 10^5.

The second line of each test case contains n integers a1,a2,…,an (−10 ^ 9≤ai≤10 ^ 9).

Output
For each test case, print the rearranged version of array a which satisfies given condition. If there are multiple valid rearrangements, print any of them.

Example
input
2
6
5 -2 4 8 6 5
4
8 1 4 2
output
5 5 4 6 8 -2
1 2 4 8

Note
In the first test case, after given rearrangement, |a1−a2|=0≤|a2−a3|=1≤|a3−a4|=2≤|a4−a5|=2≤|a5−a6|=10. There are other possible answers like “5 4 5 6 -2 8”.

In the second test case, after given rearrangement, |a1−a2|=1≤|a2−a3|=2≤|a3−a4|=4. There are other possible answers like “2 4 8 1”.

#include<iostream>
#include<algorithm>
#include<string.h>
#include<vector>
#include<stdio.h>
#include<limits.h>
#include<queue>
#include<cmath>
#include<set>
#include<map>
#define ll long long
using namespace std;
//按照新数组各元素之间差值排序 
//先排序 排序后从两头开始取 最后倒序输出
int a[100005];
int main()
{
	int t;
	int n;
	cin >> t;
	while (t--)
	{
		memset(a, 0, 100005);
		cin >> n;
		for (int i = 1; i <= n; i++)
		{
			cin >> a[i];
		}
		sort(a + 1, a + n + 1);
		vector<int>v;
		int x = 1;
		int y = n;
		while (x <= y)
		{
			v.push_back(a[y]);
			y--;
			if (x > y)
			{
				break;
			}
			v.push_back(a[x]);
			x++;
		}
		for (int i = n - 1; i >= 0; i--)
		{
			printf("%d ", v[i]);
		}
		cout << endl;
	}
	return 0;
}