使用数组元素打印长度为L的所有排列|迭代式

给定一个唯一元素数组，我们必须使用该数组的元素找到长度L的所有排列。允许重复元素。

例子：

Input: arr = { 1, 2 }, L=3
Output:
111
211
121
221
112
212
122
222

Input: arr = { 1, 2, 3 }, L=2
Output:
11
21
31
12
22
32
13
23
33

方法：

• 为了形成具有N个元素的长度L的序列，已知该序列的第i个元素可以用N种方式填充。所以会有$N^{L}$顺序
• 我们将运行一个从0到$(N^{L} - 1)$ ，对于每一个i，我们都会将i从10以基数转换为N。转换后的数字将代表数组的索引
• 我们可以打印所有$N^{L}$通过这种方式进行排序。

下面是该方法的实现：

C++

// C++ implementation
#include 
using namespace std;
  
// Convert the number to Lth
// base and print the sequence
void convert_To_Len_th_base(int n,
                            int arr[],
                            int len,
                            int L)
{
    // Sequence is of length L
    for (int i = 0; i < L; i++) {
        // Print the ith element
        // of sequence
        cout << arr[n % len];
        n /= len;
    }
    cout << endl;
}
  
// Print all the permuataions
void print(int arr[],
           int len,
           int L)
{
    // There can be (len)^l
    // permutations
    for (int i = 0; i < (int)pow(len, L); i++) {
        // Convert i to len th base
        convert_To_Len_th_base(i, arr, len, L);
    }
}
  
// Driver code
int main()
{
    int arr[] = { 1, 2, 3 };
    int len = sizeof(arr) / sizeof(arr[0]);
    int L = 2;
  
    // function call
    print(arr, len, L);
  
    return 0;
}

Java

// Java implementation for above approach
import java.io.*;
  
class GFG 
{
      
// Convert the number to Lth
// base and print the sequence
static void convert_To_Len_th_base(int n, int arr[], 
                                   int len, int L)
{
    // Sequence is of length L
    for (int i = 0; i < L; i++) 
    {
        // Print the ith element
        // of sequence
        System.out.print(arr[n % len]);
        n /= len;
    }
    System.out.println();
}
  
// Print all the permuataions
static void print(int arr[], int len, int L)
{
    // There can be (len)^l
    // permutations
    for (int i = 0; 
             i < (int)Math.pow(len, L); i++) 
    {
        // Convert i to len th base
        convert_To_Len_th_base(i, arr, len, L);
    }
}
  
// Driver code
public static void main (String[] args) 
{
    int arr[] = { 1, 2, 3 };
    int len = arr.length;
    int L = 2;
      
    // function call
    print(arr, len, L);
}
}
  
// This code is contributed by ajit.

Python3

# Python3 implementation for the above approach
  
# Convert the number to Lth
# base and print the sequence
def convert_To_Len_th_base(n, arr, Len, L):
      
    # Sequence is of Length L
    for i in range(L):
          
        # Print the ith element
        # of sequence
        print(arr[n % Len], end = "")
        n //= Len
    print()
  
# Print all the permuataions
def printf(arr, Len, L):
      
    # There can be (Len)^l permutations
    for i in range(pow(Len, L)):
          
        # Convert i to Len th base
        convert_To_Len_th_base(i, arr, Len, L)
  
# Driver code
arr = [1, 2, 3]
Len = len(arr)
L = 2
  
# function call
printf(arr, Len, L)
  
# This code is contributed by Mohit Kumar

C#

// C# implementation for above approach
using System;
  
class GFG 
{
      
// Convert the number to Lth
// base and print the sequence
static void convert_To_Len_th_base(int n, int []arr, 
                                   int len, int L)
{
    // Sequence is of length L
    for (int i = 0; i < L; i++) 
    {
        // Print the ith element
        // of sequence
        Console.Write(arr[n % len]);
        n /= len;
    }
    Console.WriteLine();
}
  
// Print all the permuataions
static void print(int []arr, int len, int L)
{
    // There can be (len)^l
    // permutations
    for (int i = 0; 
            i < (int)Math.Pow(len, L); i++) 
    {
        // Convert i to len th base
        convert_To_Len_th_base(i, arr, len, L);
    }
}
  
// Driver code
public static void Main (String[] args) 
{
    int []arr = { 1, 2, 3 };
    int len = arr.Length;
    int L = 2;
      
    // function call
    print(arr, len, L);
}
}
  
// This code is contributed by Rajput-Ji

输出：

11
21
31
12
22
32
13
23
33