将十进制转换为二进制 Java

Question

I am trying to convert decimal to binary numbers from the user's input using Java.

I'm getting errors.

package reversedBinary;
import java.util.Scanner;

public class ReversedBinary {


public static void main(String[] args) {
    int number; 

    Scanner in = new Scanner(System.in);

    System.out.println("Enter a positive integer");
    number=in.nextInt();

    if (number <0)
        System.out.println("Error: Not a positive integer");
    else { 

        System.out.print("Convert to binary is:");
        System.out.print(binaryform(number));
}

}

private static Object binaryform(int number) {
    int remainder;

    if (number <=1) {
        System.out.print(number);

    }

    remainder= number %2; 
    binaryform(number >>1);
    System.out.print(remainder);

    { 
    return null;
} } }

How do I convert Decimal to Binary in Java?

Answer 1

Integer.toBinaryString()是一个内置的方法，会做得很好。

Answer 2

Integer.toString(n,8) // decimal to octal

Integer.toString(n,2) // decimal to binary

Integer.toString(n,16) //decimal to Hex

where n = decimal number.

Answer 3

Your binaryForm method is getting caught in an infinite recursion, you need to return if number <= 1 :

import java.util.Scanner;

public class ReversedBinary {

    public static void main(String[] args) {
        int number;

        Scanner in = new Scanner(System.in);

        System.out.println("Enter a positive integer");
        number = in.nextInt();

        if (number < 0) {
            System.out.println("Error: Not a positive integer");
        } else {

            System.out.print("Convert to binary is:");
            //System.out.print(binaryform(number));
            printBinaryform(number);
        }
    }

    private static void printBinaryform(int number) {
        int remainder;

        if (number <= 1) {
            System.out.print(number);
            return; // KICK OUT OF THE RECURSION
        }

        remainder = number % 2;
        printBinaryform(number >> 1);
        System.out.print(remainder);
    }
}

Answer 4

I just want to add, for anyone who uses:

   String x=Integer.toBinaryString()

to get a String of Binary numbers and wants to convert that string into an int. If you use

  int y=Integer.parseInt(x)

you will get a NumberFormatException error.

What I did to convert String x to Integers, was first converted each individual Char in the String x to a single Char in a for loop.

  char t = (x.charAt(z));

I then converted each Char back into an individual String,

  String u=String.valueOf(t);

then Parsed each String into an Integer.

Id figure Id post this, because I took me a while to figure out how to get a binary such as 01010101 into Integer form.

Answer 5

/**
 * @param no
 *            : Decimal no
 * @return binary as integer array
 */
public int[] convertBinary(int no) {
    int i = 0, temp[] = new int[7];
    int binary[];
    while (no > 0) {
        temp[i++] = no % 2;
        no /= 2;
    }
    binary = new int[i];
    int k = 0;
    for (int j = i - 1; j >= 0; j--) {
        binary[k++] = temp[j];
    }

    return binary;
}

Answer 6

public static void main(String h[])
{
    Scanner sc=new Scanner(System.in);
    int decimal=sc.nextInt();

    String binary="";

    if(decimal<=0)
    {
        System.out.println("Please Enter more than 0");

    }
    else
    {
        while(decimal>0)
        {

            binary=(decimal%2)+binary;
            decimal=decimal/2;

        }
        System.out.println("binary is:"+binary);

    }

}

Answer 7

The following converts decimal to Binary with Time Complexity : O(n) Linear Time and with out any java inbuilt function

private static int decimalToBinary(int N) {
    StringBuilder builder = new StringBuilder();
    int base = 2;
    while (N != 0) {
        int reminder = N % base;
        builder.append(reminder);
        N = N / base;
    }

    return Integer.parseInt(builder.reverse().toString());
}

Answer 8

If you want to reverse the calculated binary form , you can use the StringBuffer class and simply use the reverse() method . Here is a sample program that will explain its use and calculate the binary

public class Binary {

    public StringBuffer calculateBinary(int number) {
        StringBuffer sBuf = new StringBuffer();
        int temp = 0;
        while (number > 0) {
            temp = number % 2;
            sBuf.append(temp);
            number = number / 2;
        }
        return sBuf.reverse();
    }
}


public class Main {

    public static void main(String[] args) throws IOException {
        System.out.println("enter the number you want to convert");
        BufferedReader bReader = new BufferedReader(newInputStreamReader(System.in));
        int number = Integer.parseInt(bReader.readLine());

        Binary binaryObject = new Binary();
        StringBuffer result = binaryObject.calculateBinary(number);
        System.out.println(result);
    }
}

Answer 9

It might seem silly , but if u wanna try utility function

System.out.println(Integer.parseInt((Integer.toString(i,2))));

there must be some utility method to do it directly, I cant remember.

Answer 10

In C# , but it's just the same as in Java :

public static void findOnes2(int num)
{
    int count = 0;      // count 1's 
    String snum = "";   // final binary representation
    int rem = 0;        // remainder

    while (num != 0)
    {
        rem = num % 2;           // grab remainder
        snum += rem.ToString();  // build the binary rep
        num = num / 2;
        if (rem == 1)            // check if we have a 1 
            count++;             // if so add 1 to the count
    }

    char[] arr = snum.ToCharArray();
    Array.Reverse(arr);
    String snum2 = new string(arr);
    Console.WriteLine("Reporting ...");
    Console.WriteLine("The binary representation :" + snum2);
    Console.WriteLine("The number of 1's is :" + count);
}

public static void Main()
{
    findOnes2(10);
}

Answer 11

public static void main(String[] args)
{
    Scanner in =new Scanner(System.in);
    System.out.print("Put a number : ");
    int a=in.nextInt();
    StringBuffer b=new StringBuffer();
    while(a>=1)
    {
      if(a%2!=0)
      {
        b.append(1);
       }
      else if(a%2==0)
      {
         b.append(0);
      }
      a /=2;
    }
    System.out.println(b.reverse());
}

Answer 12

All your problems can be solved with a one-liner! To incorporate my solution into your project, simply remove your binaryform(int number) method, and replace System.out.print(binaryform(number)); with System.out.println(Integer.toBinaryString(number)); .

Answer 13

Binary to Decimal without using Integer.ParseInt():

import java.util.Scanner;

//convert binary to decimal number in java without using Integer.parseInt() method.

public class BinaryToDecimalWithOutParseInt {

    public static void main(String[] args) {

        Scanner input = new Scanner( System.in );
        System.out.println("Enter a binary number: ");

        int  binarynum =input.nextInt();
        int binary=binarynum;

        int decimal = 0;
        int power = 0;

        while(true){

            if(binary == 0){

                break;

            } else {

                int temp = binary%10;
                decimal += temp*Math.pow(2, power);
                binary = binary/10;
                power++;

            }
        }
        System.out.println("Binary="+binarynum+" Decimal="+decimal); ;
    }

}

Output:

Enter a binary number:

1010

Binary=1010 Decimal=10

---

Binary to Decimal using Integer.parseInt():

import java.util.Scanner;

//convert binary to decimal number in java using Integer.parseInt() method.
public class BinaryToDecimalWithParseInt {

    public static void main(String[] args) {

        Scanner input = new Scanner( System.in );

        System.out.println("Enter a binary number: ");
        String binaryString =input.nextLine();

        System.out.println("Result: "+Integer.parseInt(binaryString,2));

    }

}

Output:

Enter a binary number:

1010

Result: 10

Answer 14

A rather simple than efficient program , yet it does the job.

        Scanner sc = new Scanner(System.in);
        System.out.println("Give me my binaries");
        int str = sc.nextInt(2);
        System.out.println(str);

Answer 15

public static String convertToBinary(int dec)
{
    String str = "";
    while(dec!=0)
    {
        str += Integer.toString(dec%2);
        dec /= 2;
    }
    return new StringBuffer(str).reverse().toString();
}

Answer 16

/**
 * converting decimal to binary
 *
 * @param n the number
 */
private static void toBinary(int n) {
    if (n == 0) {
        return; //end of recursion
    } else {
        toBinary(n / 2);
        System.out.print(n % 2);
    }
}

/**
 * converting decimal to binary string
 *
 * @param n the number
 * @return the binary string of n
 */
private static String toBinaryString(int n) {
    Stack<Integer> bits = new Stack<>();
    do {
        bits.push(n % 2);
        n /= 2;
    } while (n != 0);

    StringBuilder builder = new StringBuilder();
    while (!bits.isEmpty()) {
        builder.append(bits.pop());
    }
    return builder.toString();
}

Or you can use Integer.toString(int i, int radix)

eg: (Convert 12 to binary)

Integer.toString(12, 2)

Answer 17

Practically you can write it as a recursive function. Each function call returns their results and add to the tail of the previous result. It is possible to write this method by using java as simple as you can find below:

public class Solution {

    private static String convertDecimalToBinary(int n) {
        String output = "";
        if (n >= 1) {
            output = convertDecimalToBinary(n >> 1) + (n % 2);
        }

        return output;
    }

    public static void main(String[] args) {
        int num = 125;
        String binaryStr = convertDecimalToBinary(num);

        System.out.println(binaryStr);
    }

}

Let us take a look how is the above recursion working:

After calling convertDecimalToBinary method once, it calls itself till the value of the number will be lesser than 1 and return all of the concatenated results to the place where it called first.

References:

Java - Bitwise and Bit Shift Operators https://docs.oracle.com/javase/tutorial/java/nutsandbolts/op3.html

Answer 18

The better way of doing it:

public static void main(String [] args) throws IOException {
        BufferedReader bf = new BufferedReader(new InputStreamReader(System.in));
        int t = Integer.parseInt(bf.readLine().trim());
        double ans = 0;
        int i=0;

        while(t!=0){
           int digit = t & 1;
           ans = ans + (digit*Math.pow(10,i));
           i++;
           t =t>>1;
        }
        System.out.println((int)ans);
    }

Answer 19

public class BinaryConvert{ 

    public static void main(String[] args){
        System.out.println("Binary Result: "+ doBin(45));
    }

    static String doBin(int n){
        int b = 2;
        String r = "";
        String c = "";

        do{
            c += (n % b);
            n /= b;         
        }while(n != 0);

        for(int i = (c.length() - 1); i >=0; i--){
            r += c.charAt(i);
        }

        return r;
    }
}

Answer 20

I just solved this myself, and I wanted to share my answer because it includes the binary reversal and then conversion to decimal. I'm not a very experienced coder but hopefully this will be helpful to someone else.

What I did was push the binary data onto a stack as I was converting it, and then popped it off to reverse it and convert it back to decimal.

import java.util.Scanner;
import java.util.Stack;

public class ReversedBinary 
{
    private Stack<Integer> st;

    public ReversedBinary()
    {
        st = new Stack<>();
    }

    private int decimaltoBinary(int dec)
    {
        if(dec == 0 || dec == 1)
        {
            st.push(dec % 2);
            return dec;
        }

        st.push(dec % 2);

        dec = decimaltoBinary(dec / 2);        

        return dec;
    }

    private int reversedtoDecimal()
    {
        int revDec = st.pop();
        int i = 1;

        while(!st.isEmpty())
        {
            revDec += st.pop() * Math.pow(2, i++);
        }

        return revDec;
    }

    public static void main(String[] args)
    {
        ReversedBinary rev = new ReversedBinary();

        System.out.println("Please enter a positive integer:");

        Scanner sc = new Scanner(System.in);
        while(sc.hasNextLine())
        {
            int input = Integer.parseInt(sc.nextLine());
            if(input < 1 || input > 1000000000)
            {
                System.out.println("Integer must be between 1 and 1000000000!");
            }
            else
            {
                rev.decimaltoBinary(input);
                System.out.println("Binary to reversed, converted to decimal: " + rev.reversedtoDecimal());
            }
        }

    }
}

Answer 21

You can use the concept of Wrapper Classes to directly convert a decimal to binary,hexadecimal and octal.Below is a very simple program to convert decimal to reverse binary .Hope it contributes to your java knowledge

public class decimalToBinary
{
   public static void main(String[] args)
   {
       int a=43;//input
       String string=Integer.toBinaryString(a);  //decimal to binary(string)
       StringBuffer buffer = new StringBuffer(string);  //string to binary
       buffer.reverse(); //reverse of string buffer
       System.out.println(buffer);  //output as string

    }
}   

Answer 22

import java.util.*;

public class BinaryNumber 
{
    public static void main(String[] args)
    {
        Scanner scan = new Scanner(System.in);
        System.out.println("Enter the number");
        int n = scan.nextInt();
        int rem;
        int num =n; 
        String str="";
        while(num>0)
        {
            rem = num%2;
            str = rem + str;
            num=num/2;
        }
        System.out.println("the bunary number for "+n+" is : "+str);
    }
}

Answer 23

This is a very basic procedure, I got this after putting a general procedure on paper.

import java.util.Scanner;

    public class DecimalToBinary {

        public static void main(String[] args) {
            Scanner input = new Scanner(System.in);
            System.out.println("Enter a Number:");
            int number = input.nextInt();
            while(number!=0)
            {
                if(number%2==0) 
                {
                    number/=2;
                    System.out.print(0);//Example: 10/2 = 5     -> 0
                }
                else if(number%2==1) 
                {
                    number/=2;
                    System.out.print(1);// 5/2 = 2              -> 1
                }
                else if(number==2)
                {
                    number/=2;
                    System.out.print(01);// 2/2 = 0             -> 01   ->0101
                }
            }
        }
    }

Answer 24

//converts decimal to binary string
String convertToBinary(int decimalNumber){  
    String binary="";
    while(decimalNumber>0){
        int remainder=decimalNumber%2;
        //line below ensures the remainders are reversed
        binary=remainder+binary;
        decimalNumber=decimalNumber/2;
    }
    return binary;

}

Answer 25

One of the fastest solutions:

public static long getBinary(int n)
    {
        long res=0;
        int t=0;
        while(n>1)
        {
            t= (int) (Math.log(n)/Math.log(2));
            res = res+(long)(Math.pow(10, t));
            n-=Math.pow(2, t);
        }
        return res;
    }

Answer 26

Even better with StringBuilder using insert() in front of the decimal string under construction, without calling reverse(),

static String toBinary(int n) {
    if (n == 0) {
        return "0";
    }

    StringBuilder bldr = new StringBuilder();
    while (n > 0) {
        bldr = bldr.insert(0, n % 2);
        n = n / 2;
    }

    return bldr.toString();
}

Answer 27

Well, you can use while loop, like this,

import java.util.*;
public class DecimalToBinaryDemo
{
    // this function converts decimal to binary
    static void toBinary(int num)
    {
       // here we are storing binary number
       int binaryNumber[] = new int[1000];
       // "count" variable is counter for binary array
       int count = 0;
       while(num > 0)
       {
          // storing remainder in binary array
          binaryNumber[count] = num % 2;
          num = num / 2;
          count++;
       }
       // here we are printing binary in reverse order
       for(int a = count - 1; a >= 0; a--)
          System.out.print(binaryNumber[a]);
    }
    public static void main(String[] args)
    {
       int number = 20;
       toBinary(number);
   }
}

Output: 10100

Answer 28

No need of any java in-built functions. Simple recursion will do.

public class DecimaltoBinaryTest {
     public static void main(String[] args) {
        DecimaltoBinary decimaltoBinary = new DecimaltoBinary();
        System.out.println("hello " + decimaltoBinary.convertToBinary(1000,0));
    }

}

class DecimaltoBinary {

    public DecimaltoBinary() {
    }

    public int convertToBinary(int num,int binary) {
        if (num == 0 || num == 1) {
            return num;
        } 
        binary = convertToBinary(num / 2, binary);
        binary = binary * 10 + (num % 2);
        return binary;
    }
}

Answer 29

Here is the conversion of Decimal to Binary in three different ways

import java.util.Scanner;
public static Scanner scan = new Scanner(System.in);

    public static void conversionLogical(int ip){           ////////////My Method One 
        String str="";
        do{
            str=ip%2+str;
            ip=ip/2;

        }while(ip!=1);
        System.out.print(1+str);

    }
    public static void byMethod(int ip){                /////////////Online Method
        //Integer ii=new Integer(ip);
        System.out.print(Integer.toBinaryString(ip));
    }
    public static String recursion(int ip){             ////////////Using My Recursion

        if(ip==1)
            return "1";
        return (DecToBin.recursion(ip/2)+(ip%2));


    }

    public static void main(String[] args) {            ///Main Method

        int ip;         
        System.out.println("Enter Positive Integer");
        ip = scan.nextInt();

        System.out.print("\nResult 1 = ");  
        DecToBin.conversionLogical(ip);
        System.out.print("\nResult 2 = ");
        DecToBin.byMethod(ip);
        System.out.println("\nResult 3 = "+DecToBin.recursion(ip));
    }
}

Answer 30

    int n = 13;
    String binary = "";

    //decimal to binary
    while (n > 0) {
        int d = n & 1;
        binary = d + binary;
        n = n >> 1;
    }
    System.out.println(binary);

    //binary to decimal
    int power = 1;
    n = 0;
    for (int i = binary.length() - 1; i >= 0; i--) {
        n = n + Character.getNumericValue(binary.charAt(i)) * power;
        power = power * 2;
    }

    System.out.println(n);