找出前N个自然数的良好排列

给定整数N ，任务是打印前N个自然数的良好排列。让我们将置换的第_i^个元素表示为p _i 。
好的排列是这样的排列：对于所有1≤i≤N ，以下等式成立，

• p _pi =我
• _我！=我

基本上，以上表达式表示没有值等于其位置。
如果不存在这样好的排列，则打印-1 。
例子：

Input: N = 1 
Output: -1 
No good permutation exists.
Input: N = 2 
Output: 2 1 
Position of 2 is 1 and position of 1 is 2. 
 

方法：考虑置换p ，使p _i = i 。实际上， p是一个从1到N的数字序列，并且p _pi = i 。
现在唯一的技巧是改变排列以满足第二个等式，即p _i ！= i 。让我们交换每两个连续的元素。更正式地讲，对于每个k ： 2k≤n，让我们交换p _{2k – 1}和p _2k 。很容易看出，获得的置换满足每个n的两个方程，唯一的例外是：当n为奇数时，没有答案，我们应该打印-1 。
下面是上述方法的实现：

C++

// C++ implementation of the approach
#include 
using namespace std;
 
// Function to print the good permutation
// of first N natural numbers
int printPermutation(int n)
{
    // If n is odd
    if (n % 2 != 0)
        cout << -1;
 
    // Otherwise
    else
        for (int i = 1; i <= n / 2; i++)
            cout << 2 * i << " " << 2 * i - 1 << " ";
}
 
// Driver code
int main()
{
    int n = 4;
    printPermutation(n);
 
    return 0;
}

Java

// Java implementation of the approach
 
class GFG
{
 
// Function to print the good permutation
// of first N natural numbers
static int printPermutation(int n)
{
    // If n is odd
    if (n % 2 != 0)
    {
        System.out.println("-1");
    }
 
    // Otherwise
    else
        for (int i = 1; i <= n / 2; i++)
        {
            System.out.print(2 * i + " " + ((2 * i) - 1) + " ");
        }
         
    return n;
     
}
 
// Driver code
public static void main(String []args)
{
    int n = 4;
    printPermutation(n);
}
}
 
// This code contributed by Rajput-Ji

Python3

# Python3 implementation of the approach
 
# Function to print the good permutation
# of first N natural numbers
def printPermutation(n):
     
    # If n is odd
    if (n % 2 != 0):
        print(-1);
 
    # Otherwise
    else:
        for i in range(1, (n // 2) + 1):
            print((2 * i), (2 * i - 1), end = " ");
 
# Driver code
n = 4;
printPermutation(n);
 
# This code is contributed by mits

C#

// C# implementation of the approach
using System;
 
class GFG {
 
// Function to print the good permutation
// of first N natural numbers
static int printPermutation(int n)
{
    // If n is odd
    if (n % 2 != 0)
    {
        Console.WriteLine("-1");
    }
 
    // Otherwise
    else
        for (int i = 1; i <= n / 2; i++)
        {
            Console.WriteLine(2 * i + " " + ((2 * i) - 1) + " ");
        }
         
    return n;
     
}
 
// Driver code
public static void Main(String []args)
{
    int n = 4;
    printPermutation(n);
}
}
 
// This code is contributed by Kirti_Mangal

Javascript

 
// Javascript implementation of the approach
 
// Function to print the good permutation
// of first N natural numbers
function printPermutation(n)
{
    // If n is odd
    if (n % 2 != 0)
        document.write(-1);
 
    // Otherwise
    else
        for (let i = 1; i <= Math.floor(n / 2); i++)
            document.write((2 * i) + " " + ((2 * i) - 1) + " ");
}
 
// Driver code
    let n = 4;
    printPermutation(n);
     
//This code is contributed by Manoj

输出：

2 1 4 3