使用数组元素打印长度 L 的所有排列|迭代

原文:https://www . geesforgeks . org/print-all-of-length-l-use-elements-of-array-iterative/

给定一个由个唯一元素组成的数组，我们必须使用数组的元素找到所有长度的排列 L 。允许重复元素。 例:

输入: arr = { 1，2 }，L=3 输出: 111 211 121 221 112 212 122 222 输入: arr = { 1，2，3 }，L=2 输出: 11

进场:

  • 为了用 N 个元素形成长度为 L 的序列，已知序列的第I个元素可以用 N 种方式填充。所以会有N^{L} 序列
  • 我们将运行一个从 0 到(N^{L} - 1) 的循环，对于每个 i ，我们将把 i 从基地 10 转换为基地 N 。转换后的数字将代表数组的索引
  • 我们可以通过这种方式打印所有的N^{L} 序列。

以下是实施办法:

C++

// C++ implementation
#include <bits/stdc++.h>
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 语言(一种计算机语言，尤用于创建网站)

// 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.

Python 3

# 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

java 描述语言

<script>
// Javascript implementation

// Convert the number to Lth
// base and print the sequence
function convert_To_Len_th_base( n, arr, len, L)
{
    // Sequence is of length L
    for ( i = 0; i < L; i++) {
        // Print the ith element
        // of sequence
        document.write(parseInt(arr[n % len]));
        n = parseInt(n / len);
    }
   document.write("<br>");
}

// Print all the permuataions
function print( arr, len, L)
{
    // There can be (len)^l
    // permutations
    for (var i = 0; i < parseInt(Math.pow(len, L)); i++) {
        // Convert i to len th base
        convert_To_Len_th_base(i, arr, len, L);
    }
}

var arr = [ 1, 2, 3 ];
var len = arr.length;
var L = 2;

// function call
print(arr, len, L);

//This code is contributed by SoumikModnal
</script>

Output: 

11
21
31
12
22
32
13
23
33

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