重新排列数组，使偶数位置大于奇数

给定一个由n个元素组成的数组A，请根据以下关系对数组进行排序：

$A[i] >= A[i-1]$

，如果我是偶数。

$A[i] <= A[i-1]$

，如果我很奇怪。
打印结果数组。
例子 ：

Input : A[] = {1, 2, 2, 1}
Output :  1 2 1 2
Explanation : 
For 1st element, 1  1, i = 2 is even.
3rd element, 1  1, i = 4 is even.

Input : A[] = {1, 3, 2}
Output : 1 3 2
Explanation : 
Here, the array is also sorted as per the conditions. 
1  1 and 2 < 3.

方法1 –
观察到该数组由[n / 2]个均匀放置的元素组成。如果我们将最大的[n / 2]个元素分配给偶数位置，将其余的元素分配给奇数位置，那么我们的问题就解决了。因为处于奇数位置的元素将始终小于处于偶数位置的元素，因为它是最大元素，反之亦然。对数组进行排序，并在偶数位置分配前[n / 2]个元素。
下面是上述方法的实现：

C++

// C++ program to rearrange the elements
// in array such that even positioned are
// greater than odd positioned elements
#include 
using namespace std;
 
void assign(int a[], int n)
{
    // Sort the array
    sort(a, a + n);
 
    int ans[n];
    int p = 0, q = n - 1;
    for (int i = 0; i < n; i++) {
        // Assign even indexes with maximum elements
        if ((i + 1) % 2 == 0)
            ans[i] = a[q--];
 
        // Assign odd indexes with remaining elements
        else
            ans[i] = a[p++];
    }
 
    // Print result
    for (int i = 0; i < n; i++)
        cout << ans[i] << " ";
}
 
// Driver Code
int main()
{
    int A[] = { 1, 3, 2, 2, 5 };
    int n = sizeof(A) / sizeof(A[0]);
    assign(A, n);
    return 0;
}

Java

// Java program to rearrange the elements
// in array such that even positioned are
// greater than odd positioned elements
import java.io.*;
import java.util.*;
 
class GFG {
 
    static void assign(int a[], int n)
    {
 
        // Sort the array
        Arrays.sort(a);
 
        int ans[] = new int[n];
        int p = 0, q = n - 1;
        for (int i = 0; i < n; i++) {
 
            // Assign even indexes with maximum elements
            if ((i + 1) % 2 == 0)
                ans[i] = a[q--];
 
            // Assign odd indexes with remaining elements
            else
                ans[i] = a[p++];
        }
 
        // Print result
        for (int i = 0; i < n; i++)
            System.out.print(ans[i] + " ");
    }
 
    // Driver code
    public static void main(String args[])
    {
        int A[] = { 1, 3, 2, 2, 5 };
        int n = A.length;
        assign(A, n);
    }
}
 
// This code is contributed by Nikita Tiwari.

Python3

# Python3 code to rearrange the
# elements in array such that
# even positioned are greater
# than odd positioned elements
 
def assign(a, n):
     
    # Sort the array
    a.sort()
     
    ans = [0] * n
    p = 0
    q = n - 1
    for i in range(n):
         
        # Assign even indexes with
        # maximum elements
        if (i + 1) % 2 == 0:
            ans[i] = a[q]
            q = q - 1
         
        # Assign odd indexes with
        # remaining elements
        else:
            ans[i] = a[p]
            p = p + 1
             
    # Print result
    for i in range(n):
        print(ans[i], end = " ")
 
# Driver Code
A = [ 1, 3, 2, 2, 5 ]
n = len(A)
assign(A, n)
 
# This code is contributed by "Sharad_Bhardwaj".

C#

// C# program to rearrange the elements
// in array such that even positioned are
// greater than odd positioned elements
using System;
 
class GFG {
 
    static void assign(int[] a, int n)
    {
        // Sort the array
        Array.Sort(a);
 
        int[] ans = new int[n];
        int p = 0, q = n - 1;
        for (int i = 0; i < n; i++) {
 
            // Assign even indexes with maximum elements
            if ((i + 1) % 2 == 0)
                ans[i] = a[q--];
 
            // Assign odd indexes with remaining elements
            else
                ans[i] = a[p++];
        }
 
        // Print result
        for (int i = 0; i < n; i++)
            Console.Write(ans[i] + " ");
    }
 
    // Driver code
    public static void Main()
    {
        int[] A = { 1, 3, 2, 2, 5 };
        int n = A.Length;
        assign(A, n);
    }
}
 
// This code is contributed by vt_m.

Javascript

 
// JavaScript program to rearrange the elements
// in array such that even positioned are
// greater than odd positioned elements
 
    function assign(a, n)
    {
  
        // Sort the array
        a.sort();
  
        let ans = [];
        let p = 0, q = n - 1;
        for (let i = 0; i < n; i++) {
  
            // Assign even indexes with maximum elements
            if ((i + 1) % 2 == 0)
                ans[i] = a[q--];
  
            // Assign odd indexes with remaining elements
            else
                ans[i] = a[p++];
        }
  
        // Print result
        for (let i = 0; i < n; i++)
            document.write(ans[i] + " ");
    }
 
// Driver code
 
        let A = [ 1, 3, 2, 2, 5 ];
        let n = A.length;
        assign(A, n);
                             

输出：

1 5 2 3 2

时间复杂度：O(n * log n)

辅助空间：O(n)
方法2 –
另一种方法是遍历第二个元素的数组，如果不满足条件，则将该元素与前一个元素交换。它的实现方式如下：

C++

// C++ program to rearrange the elements
// in the array such that even positioned are
// greater than odd positioned elements
#include 
using namespace std;
 
// swap two elements
void swap(int* a, int* b)
{
    int temp = *a;
    *a = *b;
    *b = temp;
}
 
void rearrange(int arr[], int n)
{
    for (int i = 1; i < n; i++) {
        // if index is even
        if (i % 2 == 0) {
            if (arr[i] > arr[i - 1])
                swap(&arr[i - 1], &arr[i]);
        }
        // if index is odd
        else {
            if (arr[i] < arr[i - 1])
                swap(&arr[i - 1], &arr[i]);
        }
    }
}
 
int main()
{
    int n = 5;
    int arr[] = { 1, 3, 2, 2, 5 };
    rearrange(arr, n);
    for (int i = 0; i < n; i++)
        cout << arr[i] << " ";
    cout << "\n";
    return 0;
}

Java

// Java program to rearrange the elements
// in the array such that even positioned are
// greater than odd positioned elements
class GFG
{
  public static void rearrange(int[] arr, int n)
  {
    for(int i = 1; i < n; i++)
    {
 
      // if index is even
      if(i % 2 == 0)
      {
        if(arr[i] > arr[i - 1])
        {
 
          // swap two elements
          int temp = arr[i];
          arr[i] = arr[i - 1];
          arr[i - 1] = temp;
        }
      }
 
      // if index is odd
      else
      {
        if (arr[i] < arr[i - 1])
        {
 
          // swap two elements
          int temp = arr[i];
          arr[i] = arr[i - 1];
          arr[i - 1] = temp;
        }
      }
    }
 
    for (int i = 0; i < n; i++)
    {
      System.out.print(arr[i] + " ");
    }
  }
 
  // Driver code
  public static void main(String []args)
  {
    int n = 5;
    int arr[] = {1, 3, 2, 2, 5};
    rearrange(arr, n);
  }
}
 
// This code is contributed by avanitrachhadiya2155

Python3

# Python3 program to rearrange
# the elements in the array
# such that even positioned are
# greater than odd positioned elements
def rearrange(arr, n):
 
    for i in range (1, n):
       
        # if index is even
        if (i % 2 == 0):
            if (arr[i] > arr[i - 1]):
                arr[i - 1], arr[i] = arr[i], arr[i - 1]
         
        # if index is odd
        else:
            if (arr[i] < arr[i - 1]):
                arr[i - 1], arr[i] = arr[i] , arr[i - 1]
 
if __name__ == "__main__":         
    n = 5
    arr = [1, 3, 2, 2, 5]
    rearrange(arr, n);
    for i in range (n):
        print (arr[i], end = " ")
    print ()
    
# This code is contributed by Chitranayal

C＃

// C# program to rearrange the elements
// in the array such that even positioned are
// greater than odd positioned elements
using System;
class GFG
{
public static void rearrange(int[] arr, int n)
{
    for(int i = 1; i < n; i++)
    {
 
      // if index is even
      if(i % 2 == 0)
      {
          if(arr[i] > arr[i - 1])
          {
 
          // swap two elements
          int temp = arr[i];
          arr[i] = arr[i - 1];
          arr[i - 1] = temp;
          }
      }
 
      // if index is odd
      else
      {
          if (arr[i] < arr[i - 1])
          {
 
            // swap two elements
            int temp = arr[i];
            arr[i] = arr[i - 1];
            arr[i - 1] = temp;
          }
      }
    }
 
  for (int i = 0; i < n; i++)
  {
    Console.Write(arr[i] + " ");
  }
}
 
// Driver code
public static void Main(String []args)
{
    int n = 5;
    int []arr = {1, 3, 2, 2, 5};
    rearrange(arr, n);
}
}
 
// This code is contributed by shivanisinghss2110

Java脚本

 
// JavaScript program to rearrange the elements
// in the array such that even positioned are
// greater than odd positioned elements
 
  function rearrange(arr, n)
  {
    for(let i = 1; i < n; i++)
    {
  
      // if index is even
      if(i % 2 == 0)
      {
        if(arr[i] > arr[i - 1])
        {
  
          // swap two elements
          let temp = arr[i];
          arr[i] = arr[i - 1];
          arr[i - 1] = temp;
        }
      }
  
      // if index is odd
      else
      {
        if (arr[i] < arr[i - 1])
        {
  
          // swap two elements
          let temp = arr[i];
          arr[i] = arr[i - 1];
          arr[i - 1] = temp;
        }
      }
    }
  
    for (let i = 0; i < n; i++)
    {
      document.write(arr[i] + " ");
    }
  }
      
 
// Driver code
         
    let n = 5;
    let arr = [1, 3, 2, 2, 5];
    rearrange(arr, n);
                   

输出：

1 3 2 5 2

时间复杂度： O(n)

辅助空间： O(1)