1.题目回顾 求abs函数的最大值，考虑贪心的思想。

2.思路

①最小的放中间，两遍放最大的(顺序无关)，然后再在两遍放次小的两个.....放到直到最后一个

②最大的放中间，两遍放最小的(顺序无关) 然后再在两遍放次大的两个.....放到直到最后一个

③最后一个元素需要考虑顺序的问题，贪心， 比如放的数是 1234 ，已经排好的顺序是 3 1 4，那么2就应该放在4的后面，不应该放在3的前面。

3.卡你的地方

①需要开long long 。

②本人不会双端队列，用的vector头插尾插

AC代码

#include<iostream>
#include<vector>
#include<algorithm>
#define ll long long
using namespace std;
ll n;
vector<ll> a, b, c;

ll absfun(vector<ll> x)
{
	ll ret = 0;
	for (ll i = 0; i < n - 1; ++i)
	{
		ret += abs(x[i] - x[i + 1]);
	}
	return ret;
}
ll Min(vector<ll> a)
{
	ll i1 = 1, i2 = 1,flag = 0;
	//先小
	b.insert(b.begin(), a[0]);
	for (ll i = 1; i < n;)
	{
		//先大
		b.insert(b.begin(), a[n - i1]);
		flag = 1;
		i++;
		i1++;
		if (i == n - 1) break;
		b.push_back(a[n - i1]);
		flag = 2;
		i1++;
		i++;
		if (i == n - 1) break;
		//后小
		b.insert(b.begin(), a[i2]);
		flag = 3;
		i2++;
		i++;
		if (i == n - 1) break;
		b.push_back(a[i2]);
		flag = 4;
		i2++;
		i++;
		if (i == n - 1) break;
	}
	if (flag == 1)
	{
		if (abs(a[n-i1] - b[0]) > abs(a[n-i1] - b[b.size() - 1]))
		{
			b.insert(b.begin(), a[n - i1]);
		}
		else
		{
			b.push_back(a[n - i1]);
		}
	}
	else if (flag == 2)
	{
		if (abs(a[i2] - b[0]) > abs(a[i2] - b[b.size() - 1]))
		{
			b.insert(b.begin(), a[i2]);
		}
		else
		{
			b.push_back(a[i2]);
		}
	}
	else if (flag == 3)
	{
		if (abs(a[i2] - b[0]) > abs(a[i2] - b[b.size() - 1]))
		{
			b.insert(b.begin(), a[i2]);
		}
		else
		{
			b.push_back(a[i2]);
		}
	}
	else if (flag == 4)
	{
		if (abs(a[n - i1] - b[0]) > abs(a[n - i1] - b[b.size() - 1]))
		{
			b.insert(b.begin(), a[n - i1]);
		}
		else
		{
			b.push_back(a[n-i1]);
		}
	}
	return absfun(b);
}

ll Max(vector<ll> a)
{
	ll i1 = 0, i2 = 2, flag = 0;
	//先大
	c.insert(c.begin(), a[n-1]);
	for (ll i = 1; i < n;)
	{
		//先小
		c.insert(c.begin(), a[i1]);
		flag = 1;
		i1++;
		i++;
		if (i == n - 1) break;
		c.push_back(a[i1]);
		flag = 2;
		i1++;
		i++;
		if (i == n - 1) break;
		//后大
		c.insert(c.begin(), a[n - i2]);
		flag = 3;
		i2++;
		i++;
		if (i == n - 1) break;
		c.push_back(a[n - i2]);
		flag = 4;
		i2++;
		i++;
		if (i == n - 1) break;
	}

	if (flag == 1)
	{
		if (abs(a[i1] - c[0]) > abs(a[i1] - c[c.size() - 1]))
		{
			c.insert(c.begin(), a[i1]);
		}
		else
		{
			c.push_back(a[i1]);
		}
	}
	else if (flag == 2)
	{
		if (abs(a[n-i2] - c[0]) > abs(a[n-i2] - c[c.size() - 1]))
		{
			c.insert(c.begin(), a[n-i2]);
		}
		else
		{
			c.push_back(a[n-i2]);
		}
	}
	else if (flag == 3)
	{
		if (abs(a[n - i2] - c[0]) > abs(a[n - i2] - c[c.size() - 1]))
		{
			c.insert(c.begin(), a[n - i2]);
		}
		else
		{
			c.push_back(a[n - i2]);
		}
	}
	else
	{
		if (abs(a[i1] - c[0]) > abs(a[i1] - c[c.size() - 1]))
		{
			c.insert(c.begin(), a[i1]);
		}
		else
		{
			c.push_back(a[i1]);
		}
	}
	return absfun(c);
}
int main()
{
	cin >> n;
	for (ll i = 0; i < n; ++i)
	{
		ll x;
		cin >> x;
		a.push_back(x);
	}
	sort(a.begin(), a.end());
	ll ans1 = Min(a);//情况1
	ll ans2 = Max(a);//情况2
	cout << (ans1 > ans2 ? ans1 : ans2) << endl;
	return 0;
}