最大化圆形阵列中连续差的总和

给定n个元素的数组。将数组视为圆形数组，即_n之后的元素为₁ 。任务是找到允许元素重新排列的连续元素之间差的最大和，即元素重新排列后找到| a ₁ – a ₂ |。 + | a ₂ – a ₃ | +……+ | a _{n – 1} – a _n | + | a _n – a ₁ |。
例子：

Input : arr[] = { 4, 2, 1, 8 }
Output : 18
Rearrange given array as : { 1, 8, 2, 4 }
Sum of difference between consecutive element
= |1 - 8| + |8 - 2| + |2 - 4| + |4 - 1|
= 7 + 6 + 2 + 3
= 18.

Input : arr[] = { 10, 12, 15 }
Output : 10

这个想法是使用贪婪方法，并尝试将具有更大差异的元素拉近。
考虑给定数组a ₁ ，a ₁ ，a ₂ ，…，a _{n – 1} ，a _n的排序排列，以使₁ 2 3 …。 n – 1 n 。
现在，要获得在连续元素之间具有最大差总和的答案，请按以下方式排列元素：
a ₁ ，a _n ，a ₂ ， _n-1 ，…。， _{n / 2} ，a _{(n / 2)+1}
我们可以观察到该安排产生了最佳答案，因为所有a ₁ ，a ₂ ，a ₃ …..，a _{(n / 2)-1} ，a _{n / 2}被减去两次，而a _{(n / 2)+ 1} ，a _{(n / 2)+2} ，a _{(n / 2)+3} ，…，a _{n – 1} ，a _n被添加两次。
注– _{(n / 2)+1}该术语仅对偶数n才考虑，因为对于奇数n，它被加一次并相减一次，因此被抵消。

C++

// C++ program to maximize the sum of difference
// between consecutive elements in circular array
#include 
using namespace std;
 
// Return the maximum Sum of difference between
// consecutive elements.
int maxSum(int arr[], int n)
{
    int sum = 0;
 
    // Sorting the array.
    sort(arr, arr + n);
 
    // Subtracting a1, a2, a3,....., a(n/2)-1, an/2
    // twice and adding a(n/2)+1, a(n/2)+2, a(n/2)+3,.
    // ...., an - 1, an twice.
    for (int i = 0; i < n/2; i++)
    {
        sum -= (2 * arr[i]);
        sum += (2 * arr[n - i - 1]);
    }
 
    return sum;
}
 
// Driver Program
int main()
{
    int arr[] = { 4, 2, 1, 8 };
    int n = sizeof(arr)/sizeof(arr[0]);
    cout << maxSum(arr, n) << endl;
    return 0;
}

Java

// Java program to maximize the sum of difference
// between consecutive elements in circular array
import java.io.*;
import java.util.Arrays;
  
class MaxSum
{
    // Return the maximum Sum of difference between
    // consecutive elements.
    static int maxSum(int arr[], int n)
    {
        int sum = 0;
  
        // Sorting the array.
        Arrays.sort(arr);
      
        // Subtracting a1, a2, a3,....., a(n/2)-1,
        // an/2 twice and adding a(n/2)+1, a(n/2)+2,
        // a(n/2)+3,....., an - 1, an twice.
        for (int i = 0; i < n/2; i++)
        {
            sum -= (2 * arr[i]);
            sum += (2 * arr[n - i - 1]);
        }
      
        return sum;
    }
 
    // Driver Program
    public static void main (String[] args)
    {
        int arr[] = { 4, 2, 1, 8 };
        int n = arr.length;
        System.out.println(maxSum(arr, n));
    }
}
/*This code is contributed by Prakriti Gupta*/

Python3

# Python3 program to maximize the sum of difference
# between consecutive elements in circular array
 
# Return the maximum Sum of difference
# between consecutive elements
def maxSum(arr, n):
    sum = 0
 
    # Sorting the array
    arr.sort()
 
    # Subtracting a1, a2, a3,....., a(n/2)-1, an/2
    # twice and adding a(n/2)+1, a(n/2)+2, a(n/2)+3,.
    # ...., an - 1, an twice.
    for i in range(0, int(n / 2)) :
        sum -= (2 * arr[i])
        sum += (2 * arr[n - i - 1])
 
    return sum
 
 
# Driver Program
arr = [4, 2, 1, 8]
n = len(arr)
print (maxSum(arr, n))
 
# This code is contributed by Shreyanshi Arun.

C#

// C# program to maximize the sum of difference
// between consecutive elements in circular array
using System;
 
class MaxSum {
     
    // Return the maximum Sum of difference
    // between consecutive elements.
    static int maxSum(int[] arr, int n)
    {
        int sum = 0;
 
        // Sorting the array.
        Array.Sort(arr);
 
        // Subtracting a1, a2, a3, ....., a(n/2)-1,
        // an/2 twice and adding a(n/2)+1, a(n/2)+2,
        // a(n/2)+3, ....., an - 1, an twice.
        for (int i = 0; i < n / 2; i++) {
            sum -= (2 * arr[i]);
            sum += (2 * arr[n - i - 1]);
        }
 
        return sum;
    }
 
    // Driver Program
    public static void Main()
    {
        int[] arr = { 4, 2, 1, 8 };
        int n = arr.Length;
        Console.WriteLine(maxSum(arr, n));
    }
}
 
//This Code is contributed by vt_m.

Javascript

 
// JavaScript program for the above approach
 
    // Return the maximum Sum of difference between
    // consecutive elements.
    function maxSum(arr, n)
    {
        let sum = 0;
  
        // Sorting the array.
        arr.sort();
      
        // Subtracting a1, a2, a3,....., a(n/2)-1,
        // an/2 twice and adding a(n/2)+1, a(n/2)+2,
        // a(n/2)+3,....., an - 1, an twice.
        for (let i = 0; i < n/2; i++)
        {
            sum -= (2 * arr[i]);
            sum += (2 * arr[n - i - 1]);
        }
      
        return sum;
    }
 
// Driver Code
    let arr = [ 4, 2, 1, 8 ];
    let n = arr.length;
    document.write(maxSum(arr, n));
 
// This code is contributed by chinmoy1997pal.

输出 ：

18