重新排列数组，使得 arr[i] >= arr[j] 如果 i 是偶数并且 arr[i]<=arr[j] 如果 i 是奇数并且 j < i

重新排列数组，使得 arr[i] >= arr[j] 如果 i 是偶数并且 arr[i]<=arr[j] 如果 i 是奇数并且 j < i

给定一个包含 n 个元素的数组。我们的任务是编写一个程序来重新排列数组，使得偶数位置的元素大于它之前的所有元素，奇数位置的元素小于它之前的所有元素。
例子：

Input : arr[] = {1, 2, 3, 4, 5, 6, 7}
Output : 4 5 3 6 2 7 1

Input : arr[] = {1, 2, 1, 4, 5, 6, 8, 8} 
Output : 4 5 2 6 1 8 1 8

解决这个问题的思路是首先创建原始数组的辅助副本，并对复制的数组进行排序。现在具有 n 个元素的数组中偶数位置的总数将是 floor(n/2)，剩余的是奇数位置的数量。现在使用排序数组以以下方式填充原始数组中的奇数和偶数位置：

• 奇数仓位总数为 n – floor(n/2)。从排序数组中的第 (n-floor(n/2)) 个位置开始，并将元素复制到排序数组的第一个位置。开始从这个位置向左遍历排序后的数组，并继续向右填充原始数组中的奇数位置。
• 从第 (n-floor(n/2)+1) 位置开始向右遍历排序数组，并从第 2 位置开始填充原始数组。
  

下面是上述想法的实现：

C++

// C++ program to rearrange the array
// as per the given condition
 
#include 
using namespace std;
 
// function to rearrange the array
void rearrangeArr(int arr[], int n)
{
    // total even positions
    int evenPos = n / 2;
 
    // total odd positions
    int oddPos = n - evenPos;
 
    int tempArr[n];
 
    // copy original array in an
    // auxiliary array
    for (int i = 0; i < n; i++)
        tempArr[i] = arr[i];
 
    // sort the auxiliary array
    sort(tempArr, tempArr + n);
 
    int j = oddPos - 1;
 
    // fill up odd position in original
    // array
    for (int i = 0; i < n; i += 2) {
        arr[i] = tempArr[j];
        j--;
    }
 
    j = oddPos;
 
    // fill up even positions in original
    // array
    for (int i = 1; i < n; i += 2) {
        arr[i] = tempArr[j];
        j++;
    }
 
    // display array
    for (int i = 0; i < n; i++)
        cout << arr[i] << " ";
}
 
// Driver code
int main()
{
    int arr[] = { 1, 2, 3, 4, 5, 6, 7 };
    int size = sizeof(arr) / sizeof(arr[0]);
    rearrangeArr(arr, size);
    return 0;
}

Java

// Java program to rearrange the array
// as per the given condition
import java.util.*;
import java.lang.*;
 
public class GfG{
    // function to rearrange the array
    public static void rearrangeArr(int arr[],
                                        int n)
    {
        // total even positions
        int evenPos = n / 2;
 
        // total odd positions
        int oddPos = n - evenPos;
 
        int[] tempArr = new int [n];
 
        // copy original array in an
        // auxiliary array
        for (int i = 0; i < n; i++)
            tempArr[i] = arr[i];
 
        // sort the auxiliary array
        Arrays.sort(tempArr);
 
        int j = oddPos - 1;
 
        // fill up odd position in
        // original array
        for (int i = 0; i < n; i += 2) {
            arr[i] = tempArr[j];
            j--;
        }
 
        j = oddPos;
 
        // fill up even positions in
        // original array
        for (int i = 1; i < n; i += 2) {
            arr[i] = tempArr[j];
            j++;
        }
 
        // display array
        for (int i = 0; i < n; i++)
            System.out.print(arr[i] + " ");
    }
     
    // Driver function
    public static void main(String argc[]){
        int[] arr = new int []{ 1, 2, 3, 4, 5,
                                        6, 7 };
        int size = 7;
        rearrangeArr(arr, size);
         
    }
}
 
/* This code is contributed by Sagar Shukla */

Python3

# Python3 code to rearrange the array
# as per the given condition
import array as a
import numpy as np
 
# function to rearrange the array
def rearrangeArr(arr, n):
     
    # total even positions
    evenPos = int(n / 2)
 
    # total odd positions
    oddPos = n - evenPos
 
    # initialising empty array in python
    tempArr = np.empty(n, dtype = object)
 
    # copy original array in an
    # auxiliary array
    for i in range(0, n):
         
        tempArr[i] = arr[i]
 
    # sort the auxiliary array
    tempArr.sort()
 
    j = oddPos - 1
 
    # fill up odd position in original
    # array
    for i in range(0, n, 2):
 
        arr[i] = tempArr[j]
        j = j - 1
     
    j = oddPos
 
    # fill up even positions in original
    # array
    for i in range(1, n, 2):
        arr[i] = tempArr[j]
        j = j + 1
     
    # display array
    for i in range(0, n):
        print (arr[i], end = ' ')
 
# Driver code
arr = a.array('i', [ 1, 2, 3, 4, 5, 6, 7 ])
rearrangeArr(arr, 7)
 
# This code is contributed by saloni1297

C#

// C# program to rearrange the array
// as per the given condition
using System;
 
public class GfG {
     
    // Function to rearrange the array
    public static void rearrangeArr(int []arr, int n)
    {
        // total even positions
        int evenPos = n / 2;
 
        // total odd positions
        int oddPos = n - evenPos;
 
        int[] tempArr = new int [n];
 
        // copy original array in an
        // auxiliary array
        for (int i = 0; i < n; i++)
            tempArr[i] = arr[i];
 
        // sort the auxiliary array
        Array.Sort(tempArr);
 
        int j = oddPos - 1;
 
        // Fill up odd position in
        // original array
        for (int i = 0; i < n; i += 2) {
            arr[i] = tempArr[j];
            j--;
        }
 
        j = oddPos;
 
        // Fill up even positions in
        // original array
        for (int i = 1; i < n; i += 2) {
            arr[i] = tempArr[j];
            j++;
        }
 
        // display array
        for (int i = 0; i < n; i++)
            Console.Write(arr[i] + " ");
    }
     
    // Driver Code
    public static void Main()
    {
        int[] arr = new int []{ 1, 2, 3, 4, 5, 6, 7 };
        int size = 7;
        rearrangeArr(arr, size);
    }
}
 
/* This code is contributed by vt_m */

Javascript

 
 
// Javascript program to rearrange the array
// as per the given condition
 
 
// function to rearrange the array
function rearrangeArr(arr, n)
{
    // total even positions
    let evenPos = Math.floor(n / 2);
 
    // total odd positions
    let oddPos = n - evenPos;
 
    let tempArr = new Array(n);
 
    // copy original array in an
    // auxiliary array
    for (let i = 0; i < n; i++)
        tempArr[i] = arr[i];
 
    // sort the auxiliary array
    tempArr.sort();
 
    let j = oddPos - 1;
 
    // fill up odd position in original
    // array
    for (let i = 0; i < n; i += 2) {
        arr[i] = tempArr[j];
        j--;
    }
 
    j = oddPos;
 
    // fill up even positions in original
    // array
    for (let i = 1; i < n; i += 2) {
        arr[i] = tempArr[j];
        j++;
    }
 
    // display array
    for (let i = 0; i < n; i++)
        document.write(arr[i] + " ");
}
 
// Driver code
 
    let arr = [ 1, 2, 3, 4, 5, 6, 7 ];
    let size = arr.length;
    rearrangeArr(arr, size);
    
 
//This code is contributed by Mayank Tyagi

---

输出：


4 5 3 6 2 7 1

时间复杂度：O(n logn)
辅助空间：O(n)

另一种方法-

我们可以通过定义两个变量 p 和 q 来遍历数组并从 last 赋值。

如果甚至有索引，那么我们将给它最大值，否则给它最小值。

p = 0 和 q = 结束；

p 将继续，q 将减小。

C++

#include 
using namespace std;
 
int main(){
    int n,i,j,p,q;
    int a[]= {1, 2, 1, 4, 5, 6, 8, 8};
    n=sizeof(a)/sizeof(a[0]);
    int b[n];
    for(i=0;i=0;i--){
            if(i%2!=0){
            a[i]=b[q];
            q--;
            }
            else{
                a[i]=b[p];
                p++;
            }
    }
    for(i=0;i

Java

import java.util.*;
 
class GFG{
 
  public static void main(String[] args)
  {
    int n, i, j, p, q;
    int a[] = {1, 2, 1, 4, 5, 6, 8, 8};
    n = a.length;
    int []b = new int[n];
    for(i = 0; i < n; i++)
      b[i] = a[i];
 
    Arrays.sort(b);
    p = 0; q = n - 1;
    for(i = n - 1; i >= 0; i--)
    {
      if(i % 2 != 0)
      {
        a[i] = b[q];
        q--;
      }
      else{
        a[i] = b[p];
        p++;
      }
    }
    for(i = 0; i < n; i++)
    {
      System.out.print(a[i]+" ");
    }
  }
}
 
// This code is contributed by gauravrajput1

Python3

if __name__ == '__main__':
    #n, i, j, p, q;
    a = [ 1, 2, 1, 4, 5, 6, 8, 8 ];
    n = len(a);
    b = [0]*n;
    for i in range(n):
        b[i] = a[i];
 
    b.sort();
    p = 0;
    q = n - 1;
    for i in range(n-1, -1,-1):
        if (i % 2 != 0):
            a[i] = b[q];
            q -= 1;
        else:
            a[i] = b[p];
            p += 1;
         
    for i in range(n):
        print(a[i], end=" ");
     
# This code is contributed by gauravrajput1

C＃

using System;
 
 
public class GFG{
 
  public static void Main(String[] args)
  {
    int n, i, j, p, q;
    int []a = {1, 2, 1, 4, 5, 6, 8, 8};
    n = a.Length;
    int []b = new int[n];
    for(i = 0; i < n; i++)
      b[i] = a[i];
 
    Array.Sort(b);
    p = 0; q = n - 1;
    for(i = n - 1; i >= 0; i--)
    {
      if(i % 2 != 0)
      {
        a[i] = b[q];
        q--;
      }
      else{
        a[i] = b[p];
        p++;
      }
    }
    for(i = 0; i < n; i++)
    {
      Console.Write(a[i]+" ");
    }
  }
}
 
// This code is contributed by gauravrajput1

Javascript

     
        var n, i, j, p, q;
        var a = [ 1, 2, 1, 4, 5, 6, 8, 8 ];
        n = a.length;
        var b = Array(n).fill(0);
        for (i = 0; i < n; i++)
            b[i] = a[i];
 
        b.sort();
        p = 0;
        q = n - 1;
        for (i = n - 1; i >= 0; i--) {
            if (i % 2 != 0) {
                a[i] = b[q];
                q--;
            } else {
                a[i] = b[p];
                p++;
            }
        }
        for (i = 0; i < n; i++) {
            document.write(a[i] + " ");
        }
 
// This code is contributed by gauravrajput1

这个算法将比前一个循环少 1 个。