排列出现在最长 N 的排列的字典序排列中间，由最大 K 的整数组成

给定两个正整数K和N ，任务是找到所有排列的中间存在的排列，最多为N ，由范围 [1, K] 中的整数组成，按字典顺序排列。

例子：

Input: K = 3, N = 2
Output: 2 1
Explanation: Lexicographical order of all possible permutations are:

1. {1}.
2. {1, 1}
3. {1, 2}
4. {1, 3}
5. {2}
6. {2, 1}
7. {2, 2}
8. {2, 3}
9. {3}
10. {3, 1}
11. {3, 2}
12. {3, 3}

Therefore, the middle lexicographically the smallest sequence is (N/2)^th(= 6^th) sequence, which is {2, 1}.

Input: K = 2, N = 4
Output: 1 2 2 2

朴素方法：解决给定问题的最简单方法是生成长度为[1, N] 的所有可能子序列，由[1, K] 中的整数组成。将这些元素存储在一个数组中。生成所有子序列后，对存储的子序列列表进行排序并打印列表的中间元素。

时间复杂度： O(N ^K )
辅助空间： O(N ^K )

有效的方法：上述方法可以通过检查K的奇偶校验K是奇数还是偶数，然后相应地找到字典序中的最小序列来优化上述方法。请按照以下步骤解决问题：

• 如果K 的值为偶数，则正好有一半的序列以整数K/2或更少开始。因此，结果序列是K/2后跟(N – 1)次出现K 。
• 如果值K为奇数，则将B视为包含(K + 1)/2次出现N次的序列。
  • 对于序列X，令f(X)是这样的，X _i的X置换为(K + 1 – X _I)的序列。
  • 唯一的例外是B 的前缀。
• 从序列B开始，并执行以下步骤(N – 1)/2次：
  • 如果当前序列的最后一个元素是1 ，则将其删除。
  • 否则，将最后一个元素减1 ，当序列包含少于N 个元素时，在序列B的末尾插入K 。
• 完成上述步骤后，打印数组B[] 中得到的序列。

下面是上述方法的实现：

C++

// C++ program for the above approach
 
#include 
using namespace std;
 
// Function that finds the middle the
// lexicographical smallest sequence
void lexiMiddleSmallest(int K, int N)
{
    // If K is even
    if (K % 2 == 0) {
 
        // First element is K/2
        cout << K / 2 << " ";
 
        // Remamining elements of the
        // sequence are all integer K
        for (int i = 0; i < N - 1; ++i) {
            cout << K << " ";
        }
        cout << "\n";
        exit(0);
    }
 
    // Stores the sequence when K
    // is odd
    vector a(N, (K + 1) / 2);
 
    // Iterate over the range [0, N/2]
    for (int i = 0; i < N / 2; ++i) {
 
        // Check if the sequence ends
        // with in 1 or not
        if (a.back() == 1) {
 
            // Remove the sequence
            // ending in 1
            a.pop_back();
        }
 
        // If it doesn't end in 1
        else {
 
            // Decrement by 1
            --a.back();
 
            // Insert K to the sequence
            // till its size is N
            while ((int)a.size() < N) {
                a.push_back(K);
            }
        }
    }
 
    // Print the sequence stored
    // in the vector
    for (auto i : a) {
        cout << i << " ";
    }
    cout << "\n";
}
 
// Driver Code
int main()
{
    int K = 2, N = 4;
    lexiMiddleSmallest(K, N);
 
    return 0;
}

Java

// Java program for the above approach
import java.io.*;
import java.util.*;
 
class GFG {
 
    // Function that finds the middle the
    // lexicographical smallest sequence
    static void lexiMiddleSmallest(int K, int N)
    {
       
        // If K is even
        if (K % 2 == 0) {
 
            // First element is K/2
            System.out.print(K / 2 + " ");
 
            // Remamining elements of the
            // sequence are all integer K
            for (int i = 0; i < N - 1; ++i) {
                System.out.print(K + " ");
            }
            System.out.println();
            return;
        }
 
        // Stores the sequence when K
        // is odd
        ArrayList a = new ArrayList();
 
        // Iterate over the range [0, N/2]
        for (int i = 0; i < N / 2; ++i) {
 
            // Check if the sequence ends
            // with in 1 or not
            if (a.get(a.size() - 1) == 1) {
 
                // Remove the sequence
                // ending in 1
                a.remove(a.size() - 1);
            }
 
            // If it doesn't end in 1
            else {
 
                // Decrement by 1
                int t = a.get(a.size() - 1) - 1;
                a.set(a.get(a.size() - 1), t);
 
                // Insert K to the sequence
                // till its size is N
                while (a.size() < N) {
                    a.add(K);
                }
            }
        }
 
        // Print the sequence stored
        // in the vector
        for (int i : a) {
            System.out.print(i + " ");
        }
        System.out.println();
    }
 
    // Driver Code
    public static void main(String[] args)
    {
        int K = 2, N = 4;
        lexiMiddleSmallest(K, N);
    }
}
 
// This code is contributed by Dharanendra L V.

Python3

# Python3 program for the above approach
 
# Function that finds the middle the
# lexicographical smallest sequence
def lexiMiddleSmallest(K, N):
    # If K is even
    if (K % 2 == 0):
 
        # First element is K/2
        print(K // 2,end=" ")
 
        # Remamining elements of the
        # sequence are all integer K
        for i in range(N - 1):
            print(K, end = " ")
        print()
        return
 
    # Stores the sequence when K
    # is odd
    a = [(K + 1) // 2]*(N)
 
    # Iterate over the range [0, N/2]
    for i in range(N//2):
        # Check if the sequence ends
        # with in 1 or not
        if (a[-1] == 1):
 
            # Remove the sequence
            # ending in 1
            del a[-1]
 
        # If it doesn't end in 1
        else:
 
            # Decrement by 1
            a[-1] -= 1
 
            # Insert K to the sequence
            # till its size is N
            while (len(a) < N):
                a.append(K)
 
    # Prthe sequence stored
    # in the vector
    for i in a:
        print(i, end = " ")
    print()
 
# Driver Code
if __name__ == '__main__':
    K, N = 2, 4
    lexiMiddleSmallest(K, N)
 
# This code is contributed by mohit kumar 29.

C#

// C# program for the above approach
using System;
using System.Collections.Generic;
class GFG {
 
    // Function that finds the middle the
    // lexicographical smallest sequence
    static void lexiMiddleSmallest(int K, int N)
    {
        // If K is even
        if (K % 2 == 0) {
 
            // First element is K/2
            Console.Write(K / 2 + " ");
 
            // Remamining elements of the
            // sequence are all integer K
            for (int i = 0; i < N - 1; ++i) {
                Console.Write(K + " ");
            }
            Console.WriteLine();
            return;
        }
 
        // Stores the sequence when K
        // is odd
        List a = new List();
 
        // Iterate over the range [0, N/2]
        for (int i = 0; i < N / 2; ++i) {
 
            // Check if the sequence ends
            // with in 1 or not
            if (a[a.Count - 1] == 1) {
 
                // Remove the sequence
                // ending in 1
                a.Remove(a.Count - 1);
            }
 
            // If it doesn't end in 1
            else {
 
                // Decrement by 1
                a[a.Count - 1] -= 1;
 
                // Insert K to the sequence
                // till its size is N
                while ((int)a.Count < N) {
                    a.Add(K);
                }
            }
        }
 
        // Print the sequence stored
        // in the vector
        foreach(int i in a) { Console.Write(i + " "); }
        Console.WriteLine();
    }
 
    // Driver Code
    public static void Main()
    {
        int K = 2, N = 4;
        lexiMiddleSmallest(K, N);
    }
}
 
// This code is contributed by ukasp.

Javascript

 
// javascript program for the above approach
 
    // Function that finds the middle the
    // lexicographical smallest sequence
    function lexiMiddleSmallest(K, N)
    {
        // If K is even
        if (K % 2 == 0) {
  
            // First element is K/2
            document.write(K / 2 + " ");
  
            // Remamining elements of the
            // sequence are all leteger K
            for (let i = 0; i < N - 1; ++i) {
                document.write(K + " ");
            }
            document.write("<br/>");
            return;
        }
  
        // Stores the sequence when K
        // is odd
        let a = [];
  
        // Iterate over the range [0, N/2]
        for (let i = 0; i < N / 2; ++i) {
  
            // Check if the sequence ends
            // with in 1 or not
            if (a[a.length - 1] == 1) {
  
                // Remove the sequence
                // ending in 1
                a.pop(a.length - 1);
            }
  
            // If it doesn't end in 1
            else {
  
                // Decrement by 1
                a[a.length - 1] -= 1;
  
                // Insert K to the sequence
                // till its size is N
                while (a.length < N) {
                    a.push(K);
                }
            }
        }
  
        // Prlet the sequence stored
        // in the vector
        for(let i in a) { document.write(i + " "); }
        document.write("<br/>");
    }
 
// Driver Code
 
    let K = 2, N = 4;
    lexiMiddleSmallest(K, N);
     
 

输出：

1 2 2 2

时间复杂度： O(N)
辅助空间： O(N)

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