重新排列数组，使相邻元素的差异按降序排列

给定一个包含 n 个整数的数组 a[]，任务是重新排列数组的元素，使相邻元素的差异按降序排列。
例子：

Input : arr[] = {1, 2, 3, 4, 5, 6} 
Output : 6 1 5 2 4 3
Explanation:
For first two elements the difference is abs(6-1)=5
For next two elements the difference is abs(1-5)=4
For next two elements the difference is abs(5-2)=3
For next two elements the difference is abs(2-4)=2
For next two elements the difference is abs(4-3)=1
Hence, difference array is 5, 4, 3, 2, 1.

Input : arr[] = {7, 10, 2, 4, 5}
Output : 10 2 7 4 5 
Explanation:
For first two elements the difference is abs(10-2)=8
For next two elements the difference is abs(2-7)=5
For next two elements the difference is abs(7-4)=3
For next two elements the difference is abs(4-5)=1
Hence, difference array is 8, 5, 3, 1.

方法：
为了解决上面提到的问题，我们必须对数组进行排序，然后按以下顺序为偶数大小的数组打印元素：

a[n], a[1], a[n-1], a[2] ….. a[(n/2)], [(n/2)+1] 

对于奇数大小的数组，我们必须按以下顺序打印数组元素：

a[n], a[1], a[n-1], a[2] ….. a[(n/2)+1]

下面是上述方法的实现：

C++

// C++ implementation to Rearrange array
// such that difference of adjacent
// elements is in descending order
 
#include 
using namespace std;
 
// Function to print array in given order
void printArray(int* a, int n)
{
    // Sort the array
    sort(a, a + n);
    int i = 0;
    int j = n - 1;
 
    // Check elements till the middle index
    while (i <= j) {
        // check if length is odd
        // print the middle index at last
        if (i == j) {
            cout << a[i] << " ";
        }
        // Print the remaining elements
        // in the described order
        else {
            cout << a[j] << " ";
            cout << a[i] << " ";
        }
        i = i + 1;
        j = j - 1;
    }
    cout << endl;
}
 
// Driver code
int main()
{
 
    // array declaration
    int arr1[] = { 1, 2, 3, 4, 5, 6 };
 
    // size of array
    int n1 = sizeof(arr1) / sizeof(arr1[0]);
 
    printArray(arr1, n1);
}

Java

// Java implementation to Rearrange array
// such that difference of adjacent
// elements is in descending order
import java.util.*;
 
class GFG {
     
// Function to print array in given order
static void printArray(int []a, int n)
{
     
    // Sort the array
    Arrays.sort(a);
     
    int i = 0;
    int j = n - 1;
     
    // Check elements till the
    // middle index
    while (i <= j)
    {
         
        // Check if length is odd print
        // the middle index at last
        if (i == j)
        {
            System.out.print(a[i] + " ");
        }
         
        // Print the remaining elements
        // in the described order
        else
        {
            System.out.print(a[j] + " ");
            System.out.print(a[i] + " ");
        }
        i = i + 1;
        j = j - 1;
    }
    System.out.println();
}
     
// Driver code
public static void main (String[] args)
{
     
    // Array declaration
    int arr1[] = { 1, 2, 3, 4, 5, 6 };
     
    // Size of array
    int n1 = arr1.length;
     
    printArray(arr1, n1);
}   
}
 
// This code is contributed by AnkitRai01

Python3

# Python3 implementation to rearrange 
# array such that difference of adjacent
# elements is in descending order
 
# Function to print array in given order
def printArray(a, n):
     
    # Sort the array
    a.sort();
     
    i = 0;
    j = n - 1;
 
    # Check elements till the middle index
    while (i <= j):
         
        # Check if length is odd print
        # the middle index at last
        if (i == j):
            print(a[i], end = " ");
             
        # Print the remaining elements
        # in the described order
        else :
            print(a[j], end = " ");
            print(a[i], end = " ");
         
        i = i + 1;
        j = j - 1;
 
    print();
 
# Driver code
if __name__ == "__main__" :
 
    # Array declaration
    arr1 = [ 1, 2, 3, 4, 5, 6 ];
 
    # Size of array
    n1 = len(arr1);
 
    printArray(arr1, n1);
 
# This code is contributed by AnkitRai01

C#

// C# implementation to rearrange array
// such that difference of adjacent
// elements is in descending order
using System;
 
class GFG {
         
// Function to print array in given order
static void printArray(int []a, int n)
{
         
    // Sort the array
    Array.Sort(a);
         
    int i = 0;
    int j = n - 1;
         
    // Check elements till the
    // middle index
    while (i <= j)
    {
             
        // Check if length is odd print
        // the middle index at last
        if (i == j)
        {
            Console.Write(a[i] + " ");
        }
             
        // Print the remaining elements
        // in the described order
        else
        {
            Console.Write(a[j] + " ");
            Console.Write(a[i] + " ");
        }
        i = i + 1;
        j = j - 1;
    }
    Console.WriteLine();
}
         
// Driver code
public static void Main (string[] args)
{
         
    // Array declaration
    int []arr1 = { 1, 2, 3, 4, 5, 6 };
         
    // Size of array
    int n1 = arr1.Length;
         
    printArray(arr1, n1);
}    
}
 
// This code is contributed by AnkitRai01

Javascript

 
      // JavaScript implementation to Rearrange array
      // such that difference of adjacent
      // elements is in descending order
 
      // Function to print array in given order
      function printArray(a, n)
      {
        // Sort the array
        a.sort(function (a, b) {
          return a - b;
        });
        var i = 0;
        var j = n - 1;
 
        // Check elements till the middle index
        while (i <= j) {
          // check if length is odd
          // print the middle index at last
          if (i == j) {
            document.write(a[i] + "  ");
          }
          // Print the remaining elements
          // in the described order
          else {
            document.write(a[j] + "  ");
            document.write(a[i] + "  ");
          }
          i = i + 1;
          j = j - 1;
        }
        document.write("<br>");
      }
 
      // Driver code
      // array declaration
      var arr1 = [1, 2, 3, 4, 5, 6];
 
      // size of array
      var n1 = arr1.length;
      printArray(arr1, n1);
       
       

输出：

6 1 5 2 4 3

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