📜  使用数组元素打印长度为L的所有排列|迭代式

📅  最后修改于: 2021-06-25 14:11:53             🧑  作者: Mango

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:

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i10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


Javascript


输出:
11
21
31
12
22
32
13
23
33

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