生成一个长度为 N 的数组，其中包含 K 个子数组作为它们自己长度的排列

给定整数N和K ，任务是生成一个长度为N的数组，其中恰好包含K个子数组作为1 到 X的排列，其中X是子数组长度。可能存在多个答案，您可以打印其中任何一个。如果无法构造数组，则打印-1 。

注意：长度为 N 的排列是从 1 到 N（含）的整数列表，其中每个元素仅出现一次。

例子：

Input: N = 5,  K = 4
Output: 1, 2, 3, 5, 4
Explanation: 4 subarrays which are a permutation of their own length are:
A[1] = {1}
A[1…2] = {1, 2}
A[1…3] = {1, 2, 3}
A[1…5] = {1, 2, 3, 5, 4}
Note that their exists no permutation of length 4 as a subarray.              

Input: N = 7, K = 3
Output: {1, 2, 7, 4, 5, 6, 3}
Explanation: 3 subarrays which are a permutation of their own length are: 
A[1] = {1}
A[1…2] = {1, 2}
A[1…7] = {1, 2, 7, 3, 4, 5, 6}
Their exists no permutations of lengths 3, 4, 5 and 6 as a subarray.

方法：该问题的解决方案基于以下观察。如果所有数字按从1 到 N的递增顺序排列，则总共有N个子数组作为它们自己长度的排列。如果任何值与最高值交换，则排列数会减少。因此，使数字与K相同，使第K 值最高，并保持其他人越来越多地排序将完成任务。如果N > 1 ，则至少有 2 个子数组是它们自己长度的排列。请按照以下步骤解决问题：

• 如果N > 1 且 K < 2或如果K = 0 ，则不可能有这样的数组。
• 生成一个长度为N的数组，该数组由从1 到 N的整数按排序顺序组成。
• 数组中的最大元素显然是最后一个元素N ，即arr[N-1] 。
• 将 arr[N-1]与arr[K-1]交换
• 数组是一种可能的答案。

插图：

For example take N = 5, K = 4

• Generate array containing 1 to N is sorted order.
  arr[] = {1, 2, 3, 4, 5}
  There are 5 subarrays that are permutations of their own length: {1}, {1, 2}, {1, 2, 3}, {1, 2, 3, 4}, {1, 2, 3, 4, 5}
• Here required K is 4. So swap the 4th element with the highest value.
  arr[] = {1, 2, 3, 5, 4}
  Now there are only K subarrays which are permutation of their own length: {1}, {1, 2}, {1, 2, 3}, {1, 2, 3, 4, 5}.
  Because maximum value arrives at Kth position and now 
  for subarrays of length K or greater (but less than N) the highest element gets included in the subarray 
  which is not part of a permutation containing elements from 1 to X (subarray length).

下面是上述方法的实现。

C++

// C++ program to implement the approach
#include 
using namespace std;
 
// Function to generate the required array
vector generateArray(int N, int K)
{
    vector arr;
     
    // If making an array is not possible
    if (K == 0 || (N > 1 && K < 2))
        return arr;
    arr.assign(N, 0);
     
    // Array of size N in sorted order
    for (int i = 1; i <= N; i++)
        arr[i - 1] = i;
 
    // Swap the maximum with the Kth element
    swap(arr[K - 1], arr[N - 1]);
    return arr;
}
 
// Driver code
int main()
{
    int N = 5, K = 4;
 
    // Function call
    vector ans = generateArray(N, K);
    if (ans.size() == 0)
        cout << "-1";
    else {
        for (int i : ans)
            cout << i << " ";
    }
    return 0;
}

Java

// Java code for the above approach
import java.util.*;
class GFG{
 
  // Function to swap two numbers
  static void swap(int m, int n)
  {
    // Swapping the values
    int temp = m;
    m = n;
    n = temp;
  }
 
  // Function to generate the required array
  static int[] generateArray(int N, int K)
  {
    int[] arr = new int[N];
 
    // If making an array is not possible
    if (K == 0 || (N > 1 && K < 2))
      return arr;
    for (int i = 0; i < N; i++) {
      arr[i] = 0;
    }
 
    // Array of size N in sorted order
    for (int i = 1; i <= N; i++)
      arr[i - 1] = i;
 
    // Swap the maximum with the Kth element
    swap(arr[K - 1], arr[N - 1]);
    return arr;
  }
 
  // Driver code
  public static void main(String[] args)
  {
 
    int N = 5, K = 4;
 
    // Function call
    int[] ans = generateArray(N, K);
    if (ans.length == 0)
      System.out.print("-1");
    else {
      for (int i = 0; i < ans.length; i++) {
        System.out.print(ans[i] + " ");
      }
    }
  }
}
 
// This code is contributed by sanjoy_62.

Python3

# Python program to implement the approach
 
# Function to generate the required array
def generateArray(N,  K):
    arr = []
 
    # If making an array is not possible
    if (K == 0 or (N > 1 and K < 2)):
        return arr
 
    for i in range(0, N):
        arr.append(0)
 
    # Array of size N in sorted order
    for i in range(1, N + 1):
        arr[i-1] = i
 
    # Swap the maximum with the Kth element
    arr[K - 1], arr[N - 1] = arr[N - 1], arr[K - 1]
    return arr
 
 
# Driver code
N = 5
K = 4
 
# Function call
ans = generateArray(N, K)
if (len(ans) == 0):
    print("-1")
else:
    for i in ans:
        print(i, end=' ')
 
        # This code is contributed by ninja_hattori.

C#

// C# program to implement the approach
using System;
class GFG {
 
  // Function to swap two numbers
  static void swap(int m, int n)
  {
    // Swapping the values
    int temp = m;
    m = n;
    n = temp;
  }
 
  // Function to generate the required array
  static int[] generateArray(int N, int K)
  {
    int[] arr = new int[N];
 
    // If making an array is not possible
    if (K == 0 || (N > 1 && K < 2))
      return arr;
    for (int i = 0; i < N; i++) {
      arr[i] = 0;
    }
 
    // Array of size N in sorted order
    for (int i = 1; i <= N; i++)
      arr[i - 1] = i;
 
    // Swap the maximum with the Kth element
    swap(arr[K - 1], arr[N - 1]);
    return arr;
  }
 
  // Driver code
  public static void Main()
  {
    int N = 5, K = 4;
 
    // Function call
    int[] ans = generateArray(N, K);
    if (ans.Length == 0)
      Console.Write("-1");
    else {
      for (int i = 0; i < ans.Length; i++) {
        Console.Write(ans[i] + " ");
      }
    }
  }
}
 
// This code is contributed by Samim Hossain Mondal.

Javascript

// JavaScript code for the above approach
 
// Function to generate the required array
function generateArray( N, K)
{
    let arr ;
     
    // If making an array is not possible
    if (K == 0 || (N > 1 && K < 2))
        return [];
    arr = new Array(N).fill(0);
     
    // Array of size N in sorted order
    for (let i = 1; i <= N; i++)
        arr[i - 1] = i;
 
    // Swap the maximum with the Kth element
    let temp = arr[K - 1];
     arr[K - 1] = arr[N-1];
     arr[N-1] = temp
    return arr;
}
 
// Driver code
    let N = 5, K = 4;
 
    // Function call
   let ans = generateArray(N, K);
    if (ans.length == 0)
        document.write("-1");
    else {
        for (let i of ans)
            document.write( i  + " ");
    }
     
       // This code is contributed by Potta Lokesh
    

输出

1 2 3 5 4 

时间复杂度： 在）
辅助空间： 在）