以 {a1, b1, a2, b2, a3, b3, ……, an, bn} 格式打乱 2n 个整数而不使用额外空间

给定一个包含2n 个元素的数组，格式为 { a1, a2, a3, a4, ….., an, b1, b2, b3, b4, …., bn }。任务是在不使用额外空间的情况下将数组洗牌为 {a1, b1, a2, b2, a3, b3, ……, an, bn }。

例子：

Input : arr[] = { 1, 2, 9, 15 }
Output : 1 9 2 15

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

方法一：蛮力
蛮力解决方案涉及两个嵌套循环，以将数组后半部分中的元素向左旋转。第一个循环运行 n 次以覆盖数组后半部分中的所有元素。第二个循环将元素向左旋转。请注意，第二个循环中的开始索引取决于我们正在旋转的元素，结束索引取决于我们需要向左移动多少个位置。

下面是这种方法的实现：

C++

// C++ Naive program to shuffle an array of size 2n
#include 
using namespace std;
 
// function to shuffle an array of size 2n
void shuffleArray(int a[], int n)
{
    // Rotate the element to the left
    for (int i = 0, q = 1, k = n; i < n; i++, k++, q++)
        for (int j = k; j > i + q; j--)
            swap(a[j - 1], a[j]);
}
 
// Driven Program
int main()
{
    int a[] = { 1, 3, 5, 7, 2, 4, 6, 8 };
    int n = sizeof(a) / sizeof(a[0]);
 
    shuffleArray(a, n / 2);
 
    for (int i = 0; i < n; i++)
        cout << a[i] << " ";
 
    return 0;
}

Java

// Java Naive program to shuffle an array of size 2n
 
import java.util.Arrays;
 
public class GFG {
    // method to shuffle an array of size 2n
    static void shuffleArray(int a[], int n)
    {
        // Rotate the element to the left
        for (int i = 0, q = 1, k = n; i < n; i++, k++, q++)
            for (int j = k; j > i + q; j--) {
                // swap a[j-1], a[j]
                int temp = a[j - 1];
                a[j - 1] = a[j];
                a[j] = temp;
            }
    }
 
    // Driver Method
    public static void main(String[] args)
    {
        int a[] = { 1, 3, 5, 7, 2, 4, 6, 8 };
 
        shuffleArray(a, a.length / 2);
 
        System.out.println(Arrays.toString(a));
    }
}

Python3

# Python3 Naive program to
# shuffle an array of size 2n
 
# Function to shuffle an array of size 2n
def shuffleArray(a, n):
 
    # Rotate the element to the left
    i, q, k = 0, 1, n
    while(i < n):    
        j = k
        while(j > i + q):
            a[j - 1], a[j] = a[j], a[j - 1]
            j -= 1
        i += 1
        k += 1
        q += 1
 
# Driver Code
a = [1, 3, 5, 7, 2, 4, 6, 8]
n = len(a)
shuffleArray(a, int(n / 2))
for i in range(0, n):
    print(a[i], end = " ")
 
# This code is contributed by Smitha Dinesh Semwal.

C#

// C# Naive program to shuffle an
// array of size 2n
using System;
 
class GFG {
    // method to shuffle an array of size 2n
    static void shuffleArray(int[] a, int n)
    {
        // Rotate the element to the left
        for (int i = 0, q = 1, k = n;
             i < n; i++, k++, q++)
            for (int j = k; j > i + q; j--) {
                // swap a[j-1], a[j]
                int temp = a[j - 1];
                a[j - 1] = a[j];
                a[j] = temp;
            }
    }
 
    // Driver Code
    public static void Main()
    {
        int[] a = { 1, 3, 5, 7, 2, 4, 6, 8 };
 
        shuffleArray(a, a.Length / 2);
        for (int i = 0; i < a.Length; i++)
            Console.Write(a[i] + " ");
    }
}
 
// This code is contributed
// by ChitraNayal

Javascript

// Javascript Naive program to shuffle an array of size 2n
     
    // method to shuffle an array of size 2n
    function shuffleArray(a,n)
    {
        // Rotate the element to the left
        for (let i = 0, q = 1, k = n; i < n; i++, k++, q++)
            for (let j = k; j > i + q; j--) {
                // swap a[j-1], a[j]
                let temp = a[j - 1];
                a[j - 1] = a[j];
                a[j] = temp;
            }
    }
     
    // Driver Method
    let a=[ 1, 3, 5, 7, 2, 4, 6, 8];
    shuffleArray(a, a.length / 2);
   
    document.write(a.join(" "));
     
    //This code is contributed by avanitrachhadiya2155
     

输出：

1 2 3 4 5 6 7 8 

时间复杂度： O(n ² )

方法二：(分而治之)
这个想法是使用分而治之的技术。将给定的数组分成两半(比如 arr1[] 和 arr2[])，并将 arr1[] 的后半元素与 arr2[] 的前半元素交换。对 arr1 和 arr2 递归执行此操作。

让我们用一个例子来解释。

1. 设数组为 a1, a2, a3, a4, b1, b2, b3, b4
2. 将数组分成两半： a1, a2, a3, a4 : b1, b2, b3, b4
3. 围绕中心交换元素：将a3、a4与b1、b2对应交换。
   你得到：a1、a2、b1、b2、a3、a4、b3、b4
4. 递归地将 a1, a2, b1, b2 溢出到 a1, a2 : b1, b2
   然后将 a3, a4, b3, b4 拆分为 a3, a4 : b3, b4。
5. 为我们得到的每个子数组交换中心周围的元素：
   a1、b1、a2、b2 和 a3、b3、a4、b4。

注意：此解决方案仅处理 n = 2 ^{i 且}i = 0, 1, 2, …等的情况。

下面是这种方法的实现：

C++

// C++ Effective  program to shuffle an array of size 2n
 
#include 
using namespace std;
 
// function to shuffle an array of size 2n
void shufleArray(int a[], int f, int l)
{
    if (f > l) {
        return;
    }
 
    // If only 2 element, return
    if (l - f == 1)
        return;
 
    // finding mid to divide the array
    int mid = (f + l) / 2;
 
    // using temp for swapping first half of second array
    int temp = mid + 1;
 
    // mmid is use for swapping second half for first array
    int mmid = (f + mid) / 2;
 
    // Swapping the element
    for (int i = mmid + 1; i <= mid; i++)
        swap(a[i], a[temp++]);
 
    // Recursively doing for first half and second half
    shufleArray(a, f, mid);
    shufleArray(a, mid + 1, l);
}
 
// Driven Program
int main()
{
    int a[] = { 1, 3, 5, 7, 2, 4, 6, 8 };
    int n = sizeof(a) / sizeof(a[0]);
 
    shufleArray(a, 0, n - 1);
 
    for (int i = 0; i < n; i++)
        cout << a[i] << " ";
 
    return 0;
}

Java

// Java Effective  program to shuffle an array of size 2n
 
import java.util.Arrays;
 
public class GFG {
    // method to shuffle an array of size 2n
    static void shufleArray(int a[], int f, int l)
    {
        if (f > l)
            return;
 
        // If only 2 element, return
        if (l - f == 1)
            return;
 
        // finding mid to divide the array
        int mid = (f + l) / 2;
 
        // using temp for swapping first half of second array
        int temp = mid + 1;
 
        // mmid is use for swapping second half for first array
        int mmid = (f + mid) / 2;
 
        // Swapping the element
        for (int i = mmid + 1; i <= mid; i++) {
            // swap a[i], a[temp++]
            int temp1 = a[i];
            a[i] = a[temp];
            a[temp++] = temp1;
        }
 
        // Recursively doing for first half and second half
        shufleArray(a, f, mid);
        shufleArray(a, mid + 1, l);
    }
 
    // Driver Method
    public static void main(String[] args)
    {
        int a[] = { 1, 3, 5, 7, 2, 4, 6, 8 };
 
        shufleArray(a, 0, a.length - 1);
 
        System.out.println(Arrays.toString(a));
    }
}

蟒蛇3

# Python3 effective program to
# shuffle an array of size 2n
 
# Function to shuffle an array of size 2n
def shufleArray(a, f, l):
 
    if (f > l):
        return
 
    # If only 2 element, return
    if (l - f == 1):
        return
 
    # Finding mid to divide the array
    mid = int((f + l) / 2)
 
    # Using temp for swapping first
    # half of the second array
    temp = mid + 1
 
    # Mid is use for swapping second
    # half for first array
    mmid = int((f + mid) / 2)
 
    # Swapping the element
    for i in range(mmid + 1, mid + 1):
        (a[i], a[temp]) = (a[temp], a[i])
        temp += 1
 
    # Recursively doing for first
    # half and second half
    shufleArray(a, f, mid)
    shufleArray(a, mid + 1, l)
 
 
# Driver Code
a = [1, 3, 5, 7, 2, 4, 6, 8]
n = len(a)
shufleArray(a, 0, n - 1)
 
for i in range(0, n):
    print(a[i], end = " ")
 
# This code is contributed by Smitha Dinesh Semwal

C＃

// C# program program to merge two
// sorted arrays with O(1) extra space.
using System;
 
// method to shuffle an array of size 2n
 
public class GFG {
    // method to shuffle an array of size 2n
    static void shufleArray(int[] a, int f, int l)
    {
        if (f > l)
            return;
 
        // If only 2 element, return
        if (l - f == 1)
            return;
 
        // finding mid to divide the array
        int mid = (f + l) / 2;
 
        // using temp for swapping first half of second array
        int temp = mid + 1;
 
        // mmid is use for swapping second half for first array
        int mmid = (f + mid) / 2;
 
        // Swapping the element
        for (int i = mmid + 1; i <= mid; i++) {
            // swap a[i], a[temp++]
            int temp1 = a[i];
            a[i] = a[temp];
            a[temp++] = temp1;
        }
 
        // Recursively doing for first half and second half
        shufleArray(a, f, mid);
        shufleArray(a, mid + 1, l);
    }
 
    // Driver Method
    public static void Main()
    {
        int[] a = { 1, 3, 5, 7, 2, 4, 6, 8 };
 
        shufleArray(a, 0, a.Length - 1);
        for (int i = 0; i < a.Length; i++)
            Console.Write(a[i] + " ");
    }
}
 
/*This code is contributed by 29AjayKumar*/

Javascript

 
// Javascript Effective program to
// shuffle an array of size 2n
     
// method to shuffle an array of size 2n
function shufleArray(a, f, l)
{
    if (f > l)
        return;
 
    // If only 2 element, return
    if (l - f == 1)
        return;
 
    // Finding mid to divide the array
    let mid = Math.floor((f + l) / 2);
 
    // Using temp for swapping first
    // half of second array
    let temp = mid + 1;
 
    // mmid is use for swapping second
    // half for first array
    let mmid = Math.floor((f + mid) / 2);
 
    // Swapping the element
    for(let i = mmid + 1; i <= mid; i++)
    {
         
        // Swap a[i], a[temp++]
        let temp1 = a[i];
        a[i] = a[temp];
        a[temp++] = temp1;
    }
 
    // Recursively doing for first
    // half and second half
    shufleArray(a, f, mid);
    shufleArray(a, mid + 1, l);
}
 
// Driver Code
let a = [ 1, 3, 5, 7, 2, 4, 6, 8 ];
shufleArray(a, 0, a.length - 1);
 
document.write(a.join(" "));
 
// This code is contributed by rag2127
     

输出：

1 2 3 4 5 6 7 8 

时间复杂度： O(n log n)
线性时间解

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