Codeforces Round ＃633 (Div. 2) B. Sorted Adjacent Differences

Topic link ：https://codeforces.com/problemset/problem/1339/B
Their thinking ： thinking + greedy
I didn't turn around at first , I've been thinking about how to put equal numbers in front , After reading the boss' solution, I knew that the method was wrong from the beginning .
Make the numbers form a special arrangement , Make two adjacent numbers in the array , The difference between the former and the latter has a non decreasing trend .
Think the other way first , How to make the difference between two adjacent numbers in an array show a non increasing trend ( It is easy to understand and analyze many ) Then output it in reverse .
Non incremental situation （ The big one is ahead ） Maximum case ： Naturally, it is the difference between the maximum number and the minimum number , Because it is arranged according to the absolute value of the difference , Therefore, the latter is the minimum and the next maximum , Then there are the second small value and the second large value … Finally, the middle part （ The odd number of data is finally the middle one , An even number can be arranged according to the rule ）
thus ： For the requirements of the topic , Output in reverse . Without the transition of the previous situation, it may be difficult to understand why to output from the middle to both sides .
AC Code ：

#include<iostream>
#include<algorithm>
using namespace std;
int num[100005];
int main()
{
    
	int t;
	cin >> t;
	while (t--)
	{
    
		int n;
		cin >> n;
		for (int i = 0; i < n; i++)
			cin >> num[i];
		sort(num, num + n);
		if (n % 2 == 1)
			cout << num[n / 2] << ' ';
		for (int i = n / 2 - 1; i >= 0; i--)
			cout << num[i] << ' ' << num[n - i - 1] << ' ';
		cout << endl;
	}
}