以最大最小形式重新排列数组|设置2(O(1)多余的空间)

给定一个带正整数的排序数组，请交替排列数组，即第一个元素应该是最大值，第二个最小值，第三秒最大值，第四秒最小值等等。
例子：

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

我们在下面的文章中讨论了一个解决方案：
以最大最小形式重新排列数组|集合1 ：此处讨论的解决方案需要额外的空间，如何使用O(1)额外的空间解决此问题。

在本文中，将讨论需要O(n)时间和O(1)额外空间的解决方案。这个想法是使用乘法和模块化技巧在索引处存储两个元素。

even index : remaining maximum element.
odd index  : remaining minimum element.
 
max_index : Index of remaining maximum element
            (Moves from right to left)
min_index : Index of remaining minimum element
            (Moves from left to right)

Initialize: max_index = 'n-1'
            min_index = 0  
            max_element = arr[max_index] + 1 //can be any element which is more than the maximum value in array

For i = 0 to n-1            
    If 'i' is even
       arr[i] += arr[max_index] % max_element * max_element 
       max_index--     
    ELSE // if 'i' is odd
       arr[i] +=  arr[min_index] % max_element * max_element
       min_index++

表达式“ arr [i] + = arr [max_index]％max_element * max_element”如何工作?
该表达式的目的是将两个元素存储在索引arr [i]中。将arr [max_index]存储为乘数，将“ arr [i]”存储为余数。例如，在{1 2 3 4 5 6 7 8 9}中，max_element为10，我们将91存储在索引0中。使用91，我们可以将原始元素获取为91％10，将新元素获取为91/10。
下面实现以上思路：

C++

// C++ program to rearrange an array in minimum
// maximum form
#include 
using namespace std;
 
// Prints max at first position, min at second position
// second max at third position, second min at fourth
// position and so on.
void rearrange(int arr[], int n)
{
    // initialize index of first minimum and first
    // maximum element
    int max_idx = n - 1, min_idx = 0;
 
    // store maximum element of array
    int max_elem = arr[n - 1] + 1;
 
    // traverse array elements
    for (int i = 0; i < n; i++) {
        // at even index : we have to put maximum element
        if (i % 2 == 0) {
            arr[i] += (arr[max_idx] % max_elem) * max_elem;
            max_idx--;
        }
 
        // at odd index : we have to put minimum element
        else {
            arr[i] += (arr[min_idx] % max_elem) * max_elem;
            min_idx++;
        }
    }
 
    // array elements back to it's original form
    for (int i = 0; i < n; i++)
        arr[i] = arr[i] / max_elem;
}
 
// Driver program to test above function
int main()
{
    int arr[] = { 1, 2, 3, 4, 5, 6, 7, 8, 9 };
    int n = sizeof(arr) / sizeof(arr[0]);
 
    cout << "Original Arrayn";
    for (int i = 0; i < n; i++)
        cout << arr[i] << " ";
 
    rearrange(arr, n);
 
    cout << "\nModified Array\n";
    for (int i = 0; i < n; i++)
        cout << arr[i] << " ";
    return 0;
}

Java

// Java program to rearrange an
// array in minimum maximum form
 
public class Main {
 
    // Prints max at first position, min at second
    // position second max at third position, second
    // min at fourth position and so on.
    public static void rearrange(int arr[], int n)
    {
        // initialize index of first minimum and first
        // maximum element
        int max_idx = n - 1, min_idx = 0;
 
        // store maximum element of array
        int max_elem = arr[n - 1] + 1;
 
        // traverse array elements
        for (int i = 0; i < n; i++) {
            // at even index : we have to put
            // maximum element
            if (i % 2 == 0) {
                arr[i] += (arr[max_idx] % max_elem) * max_elem;
                max_idx--;
            }
 
            // at odd index : we have to put minimum element
            else {
                arr[i] += (arr[min_idx] % max_elem) * max_elem;
                min_idx++;
            }
        }
 
        // array elements back to it's original form
        for (int i = 0; i < n; i++)
            arr[i] = arr[i] / max_elem;
    }
 
    // Driver code
    public static void main(String args[])
    {
        int arr[] = { 1, 2, 3, 4, 5, 6, 7, 8, 9 };
        int n = arr.length;
 
        System.out.println("Original Array");
        for (int i = 0; i < n; i++)
            System.out.print(arr[i] + " ");
 
        rearrange(arr, n);
 
        System.out.print("\nModified Array\n");
        for (int i = 0; i < n; i++)
            System.out.print(arr[i] + " ");
    }
}
 
// This code is contributed by Swetank Modi

Python3

# Python3 program to rearrange an
# array in minimum maximum form
 
# Prints max at first position, min at second position
# second max at third position, second min at fourth
# position and so on.
def rearrange(arr, n):
 
    # Initialize index of first minimum
    # and first maximum element
    max_idx = n - 1
    min_idx = 0
 
    # Store maximum element of array
    max_elem = arr[n-1] + 1
 
    # Traverse array elements
    for i in range(0, n) :
 
        # At even index : we have to put maximum element
        if i % 2 == 0 :
            arr[i] += (arr[max_idx] % max_elem ) * max_elem
            max_idx -= 1
 
        # At odd index : we have to put minimum element
        else :
            arr[i] += (arr[min_idx] % max_elem ) * max_elem
            min_idx += 1
 
    # array elements back to it's original form
    for i in range(0, n) :
        arr[i] = arr[i] / max_elem
 
 
# Driver Code
arr = [1, 2, 3, 4, 5, 6, 7, 8, 9]
n = len(arr)
 
print ("Original Array")
 
for i in range(0, n):
    print (arr[i], end = " ")
     
rearrange(arr, n)
 
print ("\nModified Array")
for i in range(0, n):
    print (int(arr[i]), end = " ")
     
# This code is contributed by Shreyanshi Arun.

C#

// C# program to rearrange an
// array in minimum maximum form
using System;
 
class main {
 
    // Prints max at first position, min at second
    // position, second max at third position, second
    // min at fourth position and so on.
    public static void rearrange(int[] arr, int n)
    {
        // initialize index of first minimum
        // and first maximum element
        int max_idx = n - 1, min_idx = 0;
 
        // store maximum element of array
        int max_elem = arr[n - 1] + 1;
 
        // traverse array elements
        for (int i = 0; i < n; i++) {
 
            // at even index : we have to put
            // maximum element
            if (i % 2 == 0) {
                arr[i] += (arr[max_idx] % max_elem) * max_elem;
                max_idx--;
            }
 
            // at odd index : we have to
            // put minimum element
            else {
                arr[i] += (arr[min_idx] % max_elem) * max_elem;
                min_idx++;
            }
        }
 
        // array elements back to it's original form
        for (int i = 0; i < n; i++)
            arr[i] = arr[i] / max_elem;
    }
 
    // Driver code
    public static void Main()
    {
        int[] arr = { 1, 2, 3, 4, 5, 6, 7, 8, 9 };
        int n = arr.Length;
        Console.WriteLine("Original Array");
        for (int i = 0; i < n; i++)
            Console.Write(arr[i] + " ");
        Console.WriteLine();
 
        rearrange(arr, n);
 
        Console.WriteLine("Modified Array");
        for (int i = 0; i < n; i++)
            Console.Write(arr[i] + " ");
    }
}
 
// This code is contributed by vt_m.

---

Javascript

 
// JavaScript program to rearrange an array in minimum
// maximum form
 
// Prints max at first position, min at second position
// second max at third position, second min at fourth
// position and so on.
function rearrange(arr, n)
{
    // initialize index of first minimum and first
    // maximum element
    let max_idx = n - 1, min_idx = 0;
 
    // store maximum element of array
    let max_elem = arr[n - 1] + 1;
 
    // traverse array elements
    for (let i = 0; i < n; i++) {
        // at even index : we have to put maximum element
        if (i % 2 == 0) {
            arr[i] += (arr[max_idx] % max_elem) * max_elem;
            max_idx--;
        }
 
        // at odd index : we have to put minimum element
        else {
            arr[i] += (arr[min_idx] % max_elem) * max_elem;
            min_idx++;
        }
    }
 
    // array elements back to it's original form
    for (let i = 0; i < n; i++)
        arr[i] = Math.floor(arr[i] / max_elem);
}
 
// Driver program to test above function
 
    let arr = [ 1, 2, 3, 4, 5, 6, 7, 8, 9 ];
    let n = arr.length;
 
    document.write("Original Array<br>");
    for (let i = 0; i < n; i++)
        document.write(arr[i] + " ");
 
    rearrange(arr, n);
 
    document.write("<br>Modified Array<br>");
    for (let i = 0; i < n; i++)
        document.write(arr[i] + " ");
 
// This code is contributed by Surbhi Tyagi.
 

输出 ：

Original Array
1 2 3 4 5 6 7 8 9 
Modified Array
9 1 8 2 7 3 6 4 5 

感谢Saurabh Srivastava和Gaurav Ahirwar提出了这种方法。
另一种方法：一种更简单的方法是观察最大元素和最小元素的索引位置。偶数索引存储最大元素，奇数索引存储最小元素。随着索引的增加，最大元素减少一个，最小元素增加一个。可以完成一个简单的遍历，并再次填写arr []。
注意：仅当给定排序数组的元素是连续的，即相差一个单位时，此方法才有效。
下面是上述方法的实现：

C++

// C++ program to rearrange an array in minimum
// maximum form
#include 
using namespace std;
 
// Prints max at first position, min at second position
// second max at third position, second min at fourth
// position and so on.
void rearrange(int arr[], int n)
{
    // initialize index of first minimum and first
    // maximum element
    int max_ele = arr[n - 1];
    int min_ele = arr[0];
    // traverse array elements
    for (int i = 0; i < n; i++) {
        // at even index : we have to put maximum element
        if (i % 2 == 0) {
            arr[i] = max_ele;
            max_ele -= 1;
        }
 
        // at odd index : we have to put minimum element
        else {
            arr[i] = min_ele;
            min_ele += 1;
        }
    }
}
 
// Driver program to test above function
int main()
{
    int arr[] = { 1, 2, 3, 4, 5, 6, 7, 8, 9 };
    int n = sizeof(arr) / sizeof(arr[0]);
 
    cout << "Original Array\n";
    for (int i = 0; i < n; i++)
        cout << arr[i] << " ";
 
    rearrange(arr, n);
 
    cout << "\nModified Array\n";
    for (int i = 0; i < n; i++)
        cout << arr[i] << " ";
    return 0;
}

Java

// Java program to rearrange an
// array in minimum maximum form
 
public class Main {
 
    // Prints max at first position, min at second
    // position second max at third position, second
    // min at fourth position and so on.
    public static void rearrange(int arr[], int n)
    {
        // initialize index of first minimum and first
        // maximum element
        int max_ele = arr[n - 1];
        int min_ele = arr[0];
        // traverse array elements
        for (int i = 0; i < n; i++) {
            // at even index : we have to put maximum element
            if (i % 2 == 0) {
                arr[i] = max_ele;
                max_ele -= 1;
            }
 
            // at odd index : we have to put minimum element
            else {
                arr[i] = min_ele;
                min_ele += 1;
            }
        }
    }
 
    // Driver code
    public static void main(String args[])
    {
        int arr[] = { 1, 2, 3, 4, 5, 6, 7, 8, 9 };
        int n = arr.length;
 
        System.out.println("Original Array");
        for (int i = 0; i < n; i++)
            System.out.print(arr[i] + " ");
 
        rearrange(arr, n);
 
        System.out.print("\nModified Array\n");
        for (int i = 0; i < n; i++)
            System.out.print(arr[i] + " ");
    }
}

Python3

# Python 3 program to rearrange an
# array in minimum maximum form
 
# Prints max at first position, min
# at second position second max at
# third position, second min at
# fourth position and so on.
def rearrange(arr, n):
 
    # initialize index of first minimum
    # and first maximum element
    max_ele = arr[n - 1]
    min_ele = arr[0]
 
    # traverse array elements
    for i in range(n):
         
        # at even index : we have to
        # put maximum element
        if i % 2 == 0:
            arr[i] = max_ele
            max_ele -= 1
 
        # at odd index : we have to
        # put minimum element
        else:
            arr[i] = min_ele
            min_ele += 1
 
# Driver code
arr = [1, 2, 3, 4, 5, 6, 7, 8, 9]
n = len(arr)
print("Origianl Array")
for i in range(n):
    print(arr[i], end = " ")
 
rearrange(arr, n)
print("\nModified Array")
for i in range(n):
    print(arr[i], end = " ")
 
# This code is contributed by Shrikant13

C＃

// C# program to rearrange
// an array in minimum
// maximum form
using System;
 
class GFG
{
    // Prints max at first
    // position, min at second
    // position second max at
    // third position, second
    // min at fourth position
    // and so on.
    public static void rearrange(int []arr,
                                 int n)
    {
        // initialize index of
        // first minimum and
        // first maximum element
        int max_ele = arr[n - 1];
        int min_ele = arr[0];
         
        // traverse array elements
        for (int i = 0; i < n; i++)
        {
            // at even index : we have
            // to put maximum element
            if (i % 2 == 0)
            {
                arr[i] = max_ele;
                max_ele -= 1;
            }
 
            // at odd index : we have
            // to put minimum element
            else
            {
                arr[i] = min_ele;
                min_ele += 1;
            }
        }
    }
 
    // Driver code
    static public void Main ()
    {
        int []arr = {1, 2, 3, 4,
                     5, 6, 7, 8, 9};
        int n = arr.Length;
 
        Console.WriteLine("Original Array");
        for (int i = 0; i < n; i++)
            Console.Write(arr[i] + " ");
 
        rearrange(arr, n);
 
        Console.Write("\nModified Array\n");
        for (int i = 0; i < n; i++)
            Console.Write(arr[i] + " ");
    }
}
 
// This code is contributed by ajit

Java脚本

 
// Javascript program to rearrange an array in minimum
// maximum form
 
// Prints max at first position, min at second
    // position second max at third position, second
    // min at fourth position and so on.
    function rearrange(arr, n)
    {
        // initialize index of first minimum and first
        // maximum element
        let max_ele = arr[n - 1];
        let min_ele = arr[0];
        // traverse array elements
        for (let i = 0; i < n; i++) {
            // at even index : we have to put maximum element
            if (i % 2 == 0) {
                arr[i] = max_ele;
                max_ele -= 1;
            }
 
            // at odd index : we have to put minimum element
            else {
                arr[i] = min_ele;
                min_ele += 1;
            }
        }
    }
 
// Driver Code
 
    let arr = [ 1, 2, 3, 4, 5, 6, 7, 8, 9 ];
    let n = arr.length;
 
    document.write("Original Array" + "<br />");
    for (let i = 0; i < n; i++)
        document.write(arr[i] + " ");
    document.write("<br />");
     
    rearrange(arr, n);
 
    document.write("\nModified Array\n" + "<br />");
    for (let i = 0; i < n; i++)
       document.write(arr[i] + " ");
 

输出 ：

Original Array
1 2 3 4 5 6 7 8 9 
Modified Array
9 1 8 2 7 3 6 4 5 

感谢Apollo Doley提出了这种方法。