Dado un número entero positivo N . La tarea es generar todos los números binarios de N dígitos . Estos números binarios deben estar en orden ascendente .
Ejemplos:
Entrada: 2
Salida:
00
01
10
11
Explicación: Estos 4 son los únicos números binarios que tienen 2 dígitos.Entrada: 3
Salida:
0 0 0
0 0 1
0 1 0
0 1 1
1 0 0
1 0 1
1 1 0
1 1 1
Enfoque: Para cualquier longitud de dígito N, habrá 2 N números binarios.
- Por lo tanto, recorra de 0 a 2 N y convierta cada número a binario.
- Guarde cada número e imprímalo al final.
A continuación se muestra la implementación del enfoque anterior.
C++
// C++ code to implement above approach
#include <bits/stdc++.h>
using namespace std;
// Function to convert number
// to binary of N bits
vector<int> convertToBinary(int num,
int length)
{
// Vector to store the number
vector<int> bits(length, 0);
if (num == 0) {
return bits;
}
int i = length - 1;
while (num != 0) {
bits[i--] = (num % 2);
// Integer division
// gives quotient
num = num / 2;
}
return bits;
}
// Function to generate all
// N bit binary numbers
vector<vector<int> > getAllBinary(int n)
{
vector<vector<int> > binary_nos;
// Loop to generate the binary numbers
for (int i = 0; i < pow(2, n); i++) {
vector<int> bits =
convertToBinary(i, n);
binary_nos.push_back(bits);
}
return binary_nos;
}
// Driver code
int main()
{
int N = 3;
vector<vector<int> > binary_nos =
getAllBinary(N);
for (int i = 0; i < binary_nos.size();
i++) {
for (int j = 0;
j < binary_nos[i].size(); j++)
cout << binary_nos[i][j];
cout << endl;
}
return 0;
}
Java
// Java code for the above approach
import java.io.*;
class GFG {
// Function to convert number
// to binary of N bits
static int[] convertToBinary(int num,
int length)
{
// Vector to store the number
int[] bits = new int[length];
if (num == 0) {
return bits;
}
int i = length - 1;
while (num != 0) {
bits[i--] = (num % 2);
// Integer division
// gives quotient
num = num / 2;
}
return bits;
}
// Function to generate all
// N bit binary numbers
static int[][] getAllBinary(int n)
{
int[][] binary_nos = new int[(int)Math.pow(2,n)][];
int k = 0;
// Loop to generate the binary numbers
for (int i = 0; i < Math.pow(2, n); i++) {
int[] bits = convertToBinary(i, n);
binary_nos[k++]= bits;
}
return binary_nos;
}
// Driver code
public static void main (String[] args)
{
int N = 3;
int[][] binary_nos = getAllBinary(N);
for (int i = 0; i < binary_nos.length; i++) {
for (int j = 0; j < binary_nos[i].length; j++)
System.out.print(binary_nos[i][j]);
System.out.println();
}
}
};
// This code is contributed by Potta Lokesh
Python3
# Python 3 code to implement above approach # Function to convert number # to binary of N bits def convertToBinary(num, length): # Vector to store the number bits = [0]*(length) if (num == 0): return bits i = length - 1 while (num != 0): bits[i] = (num % 2) i -= 1 # Integer division # gives quotient num = num // 2 return bits # Function to generate all # N bit binary numbers def getAllBinary(n): binary_nos = [] # Loop to generate the binary numbers for i in range(pow(2, n)): bits = convertToBinary(i, n) binary_nos.append(bits) return binary_nos # Driver code if __name__ == "__main__": N = 3 binary_nos = getAllBinary(N) for i in range(len(binary_nos)): for j in range(len(binary_nos[i])): print(binary_nos[i][j], end="") print() # This code is contributed by ukasp.
C#
// C# code for the above approach
using System;
using System.Collections.Generic;
public class GFG
{
// Function to convert number
// to binary of N bits
static List<int> convertToBinary(int num, int length) {
// List to store the number
List<int> bits = new List<int>();
if (num == 0) {
bits.Add(0);
bits.Add(0);
bits.Add(0);
return bits;
}
int i = length - 1;
while (num != 0) {
bits.Add(num % 2);
// int division
// gives quotient
num = num / 2;
}
while(bits.Count<3)
bits.Add(0);
return bits;
}
// Function to generate all
// N bit binary numbers
static List<List<int>> getAllBinary(int n) {
List<List<int>> binary_nos = new List<List<int>>();
// Loop to generate the binary numbers
for (int i = 0; i < Math.Pow(2, n); i++) {
List<int> bits = convertToBinary(i, n);
binary_nos.Add(bits);
}
return binary_nos;
}
// Driver code
public static void Main(String[] args) {
int N = 3;
List<List<int>> binary_nos = getAllBinary(N);
foreach(var st in binary_nos){
foreach(var s in st){
Console.Write(s);
}
Console.WriteLine();
}
}
}
// This code is contributed by Rajput-Ji
Javascript
<script>
// JavaScript code to implement above approach
// Function to convert number
// to binary of N bits
const convertToBinary = (num, length) => {
// Vector to store the number
let bits = new Array(length).fill(0);
if (num == 0) {
return bits;
}
let i = length - 1;
while (num != 0) {
bits[i--] = (num % 2);
// Integer division
// gives quotient
num = parseInt(num / 2);
}
return bits;
}
// Function to generate all
// N bit binary numbers
const getAllBinary = (n) => {
let binary_nos = [];
// Loop to generate the binary numbers
for (let i = 0; i < parseInt(Math.pow(2, n)); i++) {
let bits = convertToBinary(i, n);
binary_nos.push(bits);
}
return binary_nos;
}
// Driver code
let N = 3;
let binary_nos = getAllBinary(N);
for (let i = 0; i < binary_nos.length;
i++) {
for (let j = 0;
j < binary_nos[i].length; j++)
document.write(binary_nos[i][j]);
document.write("<br/>");
}
// This code is contributed by rakeshsahni
</script>
Producción
000 001 010 011 100 101 110 111
Complejidad de Tiempo: O(2 N )
Espacio Auxiliar: O(2 N )