每个元素出现两次且外观之间的距离等于该值的组合

给定一个正数n，我们需要找到2 * n个元素的所有组合，以使从1到n的每个元素都出现两次，并且其出现之间的距离恰好等于该元素的值。

例子：

Input :  n = 3
Output : 3 1 2 1 3 2
         2 3 1 2 1 3
All elements from 1 to 3 appear
twice and distance between two
appearances is equal to value
of the element.

Input :  n = 4
Output : 4 1 3 1 2 4 3 2
         2 3 4 2 1 3 1 4

解释

我们可以使用回溯来解决这个问题。这个想法是针对第一个元素的所有可能组合，然后递归地探索其余元素，以检查它们是否会导致解决方案。如果当前配置没有解决方案，我们将回溯。注意，可以将元素k放置在输出数组i> = 0且(i + k + 1)<2 * n的位置i和(i + k + 1)的位置。
请注意，元素的组合对于n的某些值(如2、5、6等)是不可能的。

C++

// C++ program to find all combinations where every
// element appears twice and distance between
// appearances is equal to the value
#include 
using namespace std;
  
// Find all combinations that satisfies given constraints
void allCombinationsRec(vector &arr, int elem, int n)
{
    // if all elements are filled, print the solution
    if (elem > n)
    {
        for (int i : arr)
            cout << i << " ";
        cout << endl;
  
        return;
    }
  
    // try all possible combinations for element elem
    for (int i = 0; i < 2*n; i++)
    {
        // if position i and (i+elem+1) are not occupied
        // in the vector
        if (arr[i] == -1 && (i + elem + 1) < 2*n &&
                arr[i + elem + 1] == -1)
        {
            // place elem at position i and (i+elem+1)
            arr[i] = elem;
            arr[i + elem + 1] = elem;
  
            // recurse for next element
            allCombinationsRec(arr, elem + 1, n);
  
            // backtrack (remove elem from position i and (i+elem+1) )
            arr[i] = -1;
            arr[i + elem + 1] = -1;
        }
    }
}
  
void allCombinations(int n)
{
    // create a vector of double the size of given number with
    vector arr(2*n, -1);
  
    // all its elements initialized by 1
    int elem = 1;
  
    // start from element 1
    allCombinationsRec(arr, elem, n);
}
  
// Driver code
int main()
{
    // given number
    int n = 3;
    allCombinations(n);
    return 0;
}

Java

// Java program to find all combinations where every
// element appears twice and distance between
// appearances is equal to the value
  
import java.util.Vector;
  
class Test
{
    // Find all combinations that satisfies given constraints
    static void allCombinationsRec(Vector arr, int elem, int n)
    {
        // if all elements are filled, print the solution
        if (elem > n)
        {
            for (int i : arr)
                System.out.print(i + " ");
            System.out.println();
       
            return;
        }
       
        // try all possible combinations for element elem
        for (int i = 0; i < 2*n; i++)
        {
            // if position i and (i+elem+1) are not occupied
            // in the vector
            if (arr.get(i) == -1 && (i + elem + 1) < 2*n &&
                    arr.get(i + elem + 1) == -1)
            {
                // place elem at position i and (i+elem+1)
                arr.set(i, elem);
                arr.set(i + elem + 1, elem);
       
                // recurse for next element
                allCombinationsRec(arr, elem + 1, n);
       
                // backtrack (remove elem from position i and (i+elem+1) )
                arr.set(i, -1);
                arr.set(i + elem + 1, -1);
            }
        }
    }
       
    static void allCombinations(int n)
    {
          
        // create a vector of double the size of given number with
        Vector arr = new Vector<>();
          
        for (int i = 0; i < 2*n; i++) {
            arr.add(-1);
        }
          
        // all its elements initialized by 1
        int elem = 1;
       
        // start from element 1
        allCombinationsRec(arr, elem, n);
    }
      
    // Driver method
    public static void main(String[] args) 
    {
        // given number
        int n = 3;
        allCombinations(n);
    }
}

Python3

# Python3 program to find all combinations 
# where every element appears twice and distance 
# between appearances is equal to the value
  
# Find all combinations that
# satisfies given constraints
def allCombinationsRec(arr, elem, n):
  
    # if all elements are filled,
    # print the solution
    if (elem > n):
      
        for i in (arr):
            print(i, end = " ")
              
        print("")
        return
  
    # Try all possible combinations
    # for element elem
    for i in range(0, 2 * n):
      
        # if position i and (i+elem+1) are  
        # not occupied in the vector
        if (arr[i] == -1 and
           (i + elem + 1) < 2*n and
           arr[i + elem + 1] == -1):
          
            # place elem at position
            # i and (i+elem+1)
            arr[i] = elem
            arr[i + elem + 1] = elem
  
            # recurse for next element
            allCombinationsRec(arr, elem + 1, n)
  
            # backtrack (remove elem from
            # position i and (i+elem+1) )
            arr[i] = -1
            arr[i + elem + 1] = -1
          
def allCombinations(n):
  
    # create a vector of double
    # the size of given number with
    arr = [-1] * (2 * n)
  
    # all its elements initialized by 1
    elem = 1
  
    # start from element 1
    allCombinationsRec(arr, elem, n)
  
# Driver code
n = 3
allCombinations(n)
  
# This code is contributed by Smitha Dinesh Semwal.

输出：

3 1 2 1 3 2 
 2 3 1 2 1 3