我正在尝试使用Java从用户输入转换十进制数到二进制数。
我收到错误。
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;
} } }
如何在Java中将十进制转换为二进制?
答案 0 :(得分:70)
Integer.toBinaryString()
是一种内置方法,并且效果很好。
答案 1 :(得分:33)
Integer.toString(n,8) // decimal to octal
Integer.toString(n,2) // decimal to binary
Integer.toString(n,16) //decimal to Hex
其中n =十进制数。
答案 2 :(得分:12)
您的binaryForm
方法陷入了无限递归,您需要在number <= 1
时返回:
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);
}
}
答案 3 :(得分: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;
}
答案 4 :(得分:5)
我只是想为任何使用过的人添加:
String x=Integer.toBinaryString()
获取二进制数字串并希望将该字符串转换为int。如果你使用
int y=Integer.parseInt(x)
您将收到NumberFormatException错误。
我将String x转换为Integers所做的是,首先将String x中的每个Char转换为for循环中的单个Char。
char t = (x.charAt(z));
然后我将每个Char转换回单个String,
String u=String.valueOf(t);
然后将每个String解析为整数。
Id figure Id发布此信息,因为我花了一些时间来弄清楚如何将二进制文件(如01010101)转换为整数形式。
答案 5 :(得分:3)
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);
}
}
答案 6 :(得分:1)
以下将十进制转换为具有时间复杂度的二进制:O(n)线性时间并且没有任何java内置函数
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());
}
答案 7 :(得分:0)
实际上你可以把它写成一个递归函数。每个函数调用都返回它们的结果并添加到前一个结果的尾部。可以使用 java 编写此方法,如下所示:
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);
}
}
让我们看看上面的递归是如何工作的:
调用convertDecimalToBinary方法一次后,它会调用自己,直到数字的值小于1,并将所有连接的结果返回到它首先调用的地方。
参考:
Java - 按位和位移运算符 https://docs.oracle.com/javase/tutorial/java/nutsandbolts/op3.html
答案 8 :(得分:0)
/**
* 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();
}
或者您可以使用Integer.toString(int i, int radix)
例如:(将12转换为二进制)
Integer.toString(12, 2)
答案 9 :(得分:0)
public static void converToBinary(int dec)
{
String str = "";
while(dec!=0)
{
str = str + Integer.toString(dec%2);
dec = dec/2;
}
System.out.println(new StringBuffer(str).reverse().toString());
}
答案 10 :(得分:0)
一个相当简单而不是高效的程序,但它可以完成这项工作。
Scanner sc = new Scanner(System.in);
System.out.println("Give me my binaries");
int str = sc.nextInt(2);
System.out.println(str);
答案 11 :(得分:0)
二进制到十进制而不使用Integer.ParseInt():
/
<强>输出:强>
输入二进制数:
1010
二进制= 1010十进制= 10
使用Integer.parseInt()的二进制到十进制:
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); ;
}
}
<强>输出:强>
输入二进制数:
1010
结果:10
答案 12 :(得分:0)
如果要反转计算的二进制形式,可以使用StringBuffer类并只使用reverse()方法。这是一个示例程序,将解释其用法并计算二进制文件
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);
}
}
答案 13 :(得分:0)
这可能看起来很愚蠢,但如果你想尝试效用函数
System.out.println(Integer.parseInt((Integer.toString(i,2))));
必须有一些实用方法才能直接做到,我记不起来了。
答案 14 :(得分:-1)
最快的解决方案之一:
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;
}
答案 15 :(得分:-1)
在C#中,它与Java中的相同:
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);
}
答案 16 :(得分:-1)
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;
}
}
答案 17 :(得分:-1)
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());
}
答案 18 :(得分:-1)
好吧,你可以使用while循环,像这样,
import java.util.Scanner;
public class DecimalToBinaryExample
{
public static void main(String[] args)
{
int num;
int a = 0;
Scanner sc = new Scanner(System.in);
System.out.println("Please enter a decimal number : ");
num = sc.nextInt();
int binary[] = new int[100];
while(num != 0)
{
binary[a] = num % 2;
num = num / 2;
a++;
}
System.out.println("The binary value is : ");
for(int b = a - 1; b >= 0; b--)
{
System.out.println("" + binary[b]);
}
sc.close();
}
}
您可以参考以下示例获得一些好的解释,
答案 19 :(得分:-1)
使用StringBuilder在构造中的十进制字符串前面使用insert()更好,而不调用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();
}
答案 20 :(得分:-1)
你可以使用Wrapper Classes的概念直接将十进制转换为二进制,十六进制和八进制.Below是一个非常简单的程序,用于将十进制转换为反向二进制。希望它有助于你的java知识
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
}
}
答案 21 :(得分:-1)
mMask.setVisibility(View.VISIBLE); // blocks children
mMask.setVisibility(View.GONE); // children clickable
答案 22 :(得分:-1)
我自己就解决了这个问题,我想分享我的答案,因为它包含二进制反转然后转换为十进制。我不是一个非常有经验的编码员,但希望这对其他人有帮助。
我所做的是在转换时将二进制数据压入堆栈,然后将其弹出以将其反转并将其转换回十进制数。
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());
}
}
}
}
答案 23 :(得分:-1)
所有问题都可以通过单行解决!
要将我的解决方案合并到您的项目中,只需删除您的binaryform(int number)
方法,然后将System.out.print(binaryform(number));
替换为System.out.println(Integer.toBinaryString(number));
。
答案 24 :(得分:-1)
这是一个非常基本的程序,我在将一般程序写在纸上后得到了这个。
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
}
}
}
}
答案 25 :(得分:-1)
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);
}
}
答案 26 :(得分:-2)
不需要任何java内置函数。简单的递归就可以了。
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;
}
}
答案 27 :(得分:-2)
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);
答案 28 :(得分:-2)
以下是以三种不同方式将十进制转换为二进制
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));
}
}