有谁知道如何在 Java 中添加 2 个二进制数，以二进制形式输入？
例如，1010 + 10 = 1100。
【问题讨论】：
标签： java binary decimal addition
使用Integer.parseInt(String, int radix)。
public static String addBinary(){
// The two input Strings, containing the binary representation of the two values:
String input0 = "1010";
String input1 = "10";
// Use as radix 2 because it's binary
int number0 = Integer.parseInt(input0, 2);
int number1 = Integer.parseInt(input1, 2);
int sum = number0 + number1;
return Integer.toBinaryString(sum); //returns the answer as a binary value;
}
【讨论】：
Integer 是一个类，它包含一个将表示整数的字符串解析为其实整数值 (int) 的方法。你可以看看我提供给你的链接。
2,147,483,647 并且（第一个）二进制值转换为 24,847,893,154,024,981,730,169,397,005
深入了解基础知识：
public static String binaryAddition(String s1, String s2) {
if (s1 == null || s2 == null) return "";
int first = s1.length() - 1;
int second = s2.length() - 1;
StringBuilder sb = new StringBuilder();
int carry = 0;
while (first >= 0 || second >= 0) {
int sum = carry;
if (first >= 0) {
sum += s1.charAt(first) - '0';
first--;
}
if (second >= 0) {
sum += s2.charAt(second) - '0';
second--;
}
carry = sum >> 1;
sum = sum & 1;
sb.append(sum == 0 ? '0' : '1');
}
if (carry > 0)
sb.append('1');
sb.reverse();
return String.valueOf(sb);
}
【讨论】：
Martijn 绝对正确，背着完成答案
Integer.toBinaryString(sum);
会根据 OP 问题以二进制形式提供您的输出。
【讨论】：
你可以在二进制数前面加上0b来表示它是二进制的。
对于这个例子，你可以简单地做：
Integer.toString(0b1010 + 0b10, 2);
这会将两者相加为二进制，并且 Integer.toString() 以 2 作为第二个参数将其转换回二进制。
【讨论】：
Martijn 的原始解决方案不适用于大二进制数。下面的代码可以用来克服这个问题。
public String addBinary(String s1, String s2) {
StringBuilder sb = new StringBuilder();
int i = s1.length() - 1, j = s2.length() -1, carry = 0;
while (i >= 0 || j >= 0) {
int sum = carry;
if (j >= 0) sum += s2.charAt(j--) - '0';
if (i >= 0) sum += s1.charAt(i--) - '0';
sb.append(sum % 2);
carry = sum / 2;
}
if (carry != 0) sb.append(carry);
return sb.reverse().toString();
}
【讨论】：
public class BinaryArithmetic {
/*-------------------------- add ------------------------------------------------------------*/
static String add(double a, double b) {
System.out.println(a + "first val :" + b);
int a1 = (int) a;
int b1 = (int) b;
String s1 = Integer.toString(a1);
String s2 = Integer.toString(b1);
int number0 = Integer.parseInt(s1, 2);
int number1 = Integer.parseInt(s2, 2);
int sum = number0 + number1;
String s3 = Integer.toBinaryString(sum);
return s3;
}
/*-------------------------------multiply-------------------------------------------------------*/
static String multiply(double a, double b) {
System.out.println(a + "first val :" + b);
int a1 = (int) a;
int b1 = (int) b;
String s1 = Integer.toString(a1);
String s2 = Integer.toString(b1);
int number0 = Integer.parseInt(s1, 2);
int number1 = Integer.parseInt(s2, 2);
int sum = number0 * number1;
String s3 = Integer.toBinaryString(sum);
return s3;
}
/*----------------------------------------substraction----------------------------------------------*/
static String sub(double a, double b) {
System.out.println(a + "first val :" + b);
int a1 = (int) a;
int b1 = (int) b;
String s1 = Integer.toString(a1);
String s2 = Integer.toString(b1);
int number0 = Integer.parseInt(s1, 2);
int number1 = Integer.parseInt(s2, 2);
int sum = number0 - number1;
String s3 = Integer.toBinaryString(sum);
return s3;
}
/*--------------------------------------division------------------------------------------------*/
static String div(double a, double b) {
System.out.println(a + "first val :" + b);
int a1 = (int) a;
int b1 = (int) b;
String s1 = Integer.toString(a1);
String s2 = Integer.toString(b1);
int number0 = Integer.parseInt(s1, 2);
int number1 = Integer.parseInt(s2, 2);
int sum = number0 / number1;
String s3 = Integer.toBinaryString(sum);
return s3;
}
}
【讨论】：
另一个有趣但冗长的方法是将两个数字中的每一个都转换为十进制，将十进制数字相加并将获得的答案转换回二进制！
【讨论】：
Java解决方案
static String addBinary(String a, String b) {
int lenA = a.length();
int lenB = b.length();
int i = 0;
StringBuilder sb = new StringBuilder();
int rem = Math.abs(lenA-lenB);
while(rem >0){
sb.append('0');
rem--;
}
if(lenA > lenB){
sb.append(b);
b = sb.toString();
}else{
sb.append(a);
a = sb.toString();
}
sb = new StringBuilder();
char carry = '0';
i = a.length();
while(i > 0){
if(a.charAt(i-1) == b.charAt(i-1)){
sb.append(carry);
if(a.charAt(i-1) == '1'){
carry = '1';
}else{
carry = '0';
}
}else{
if(carry == '1'){
sb.append('0');
carry = '1';
}else{
carry = '0';
sb.append('1');
}
}
i--;
}
if(carry == '1'){
sb.append(carry);
}
sb.reverse();
return sb.toString();
}
【讨论】：
public String addBinary(String a, String b) {
int carry = 0;
StringBuilder sb = new StringBuilder();
for(int i = a.length() - 1, j = b.length() - 1;i >= 0 || j >= 0;i--,j--){
int sum = carry + (i >= 0 ? a.charAt(i) - '0':0) + (j >= 0 ? b.charAt(j) - '0' : 0);
sb.append(sum%2);
carry =sum / 2;
}
if(carry > 0) sb.append(carry);
sb.reverse();
return sb.toString();
}
【讨论】：
这个想法与一些答案中讨论的相同，但这个是一个更短且更容易理解的解决方案（步骤已注释）。
// Handles numbers which are way bigger.
public String addBinary(String a, String b) {
StringBuilder sb = new StringBuilder();
int i = a.length() - 1;
int j = b.length() -1;
int carry = 0;
while (i >= 0 || j >= 0) {
int sum = carry;
if (j >= 0) { sum += b.charAt(j--) - '0' };
if (i >= 0) { sum += a.charAt(i--) - '0' };
// Added number can be only 0 or 1
sb.append(sum % 2);
// Get the carry.
carry = sum / 2;
}
if (carry != 0) { sb.append(carry); }
// First reverse and then return.
return sb.reverse().toString();
}
【讨论】：
我试图让它变得简单，这是我必须处理的密码学 prj 它效率不高，但我希望它
public String binarysum(String a, String b){
int carry=0;
int maxim;
int minim;
maxim=Math.max(a.length(),b.length());
minim=Math.min(a.length(),b.length());
char smin[]=new char[minim];
char smax[]=new char[maxim];
if(a.length()==minim){
for(int i=0;i<smin.length;i++){
smin[i]=a.charAt(i);
}
for(int i=0;i<smax.length;i++){
smax[i]=b.charAt(i);
}
}
else{
for(int i=0;i<smin.length;i++){
smin[i]=b.charAt(i);
}
for(int i=0;i<smax.length;i++){
smax[i]=a.charAt(i);
}
}
char[]sum=new char[maxim];
char[] st=new char[maxim];
for(int i=0;i<st.length;i++){
st[i]='0';
}
int k=st.length-1;
for(int i=smin.length-1;i>-1;i--){
st[k]=smin[i];
k--;
}
// *************************** sum begins here
for(int i=maxim-1;i>-1;i--){
char x= smax[i];
char y= st[i];
if(x==y && x=='0'){
if(carry==0)
sum[i]='0';
else if(carry==1){
sum[i]='1';
carry=0;
}
}
else if(x==y && x=='1'){
if(carry==0){
sum[i]='0';
carry=1;
}
else if(carry==1){
sum[i]='1';
carry=1;
}
}
else if(x!=y){
if(carry==0){
sum[i]='1';
}
else if(carry==1){
sum[i]='0';
carry=1;
}
} }
String s=new String(sum);
return s;
}
【讨论】：
这是一个python版本
def binAdd(s1, s2):
if not s1 or not s2:
return ''
maxlen = max(len(s1), len(s2))
s1 = s1.zfill(maxlen)
s2 = s2.zfill(maxlen)
result = ''
carry = 0
i = maxlen - 1
while(i >= 0):
s = int(s1[i]) + int(s2[i])
if s == 2: #1+1
if carry == 0:
carry = 1
result = "%s%s" % (result, '0')
else:
result = "%s%s" % (result, '1')
elif s == 1: # 1+0
if carry == 1:
result = "%s%s" % (result, '0')
else:
result = "%s%s" % (result, '1')
else: # 0+0
if carry == 1:
result = "%s%s" % (result, '1')
carry = 0
else:
result = "%s%s" % (result, '0')
i = i - 1;
if carry>0:
result = "%s%s" % (result, '1')
return result[::-1]
【讨论】：
Java。
实际上，我已经设法在不使用stringbuilder() 函数的情况下找到了这个问题的解决方案。看看这个：
public void BinaryAddition(String s1,String s2)
{
int l1=s1.length();int c1=l1;
int l2=s2.length();int c2=l2;
int max=(int)Math.max(l1,l2);
int arr1[]=new int[max];
int arr2[]=new int[max];
int sum[]=new int[max+1];
for(int i=(arr1.length-1);i>=(max-l1);i--)
{
arr1[i]=(int)(s1.charAt(c1-1)-48);
c1--;
}
for(int i=(arr2.length-1);i>=(max-l2);i--)
{
arr2[i]=(int)(s2.charAt(c2-1)-48);
c2--;
}
for(int i=(sum.length-1);i>=1;i--)
{
sum[i]+=arr1[i-1]+arr2[i-1];
if(sum[i]==2)
{
sum[i]=0;
sum[i-1]=1;
}
else if(sum[i]==3)
{
sum[i]=1;
sum[i-1]=1;
}
}
int c=0;
for(int i=0;i<sum.length;i++)
{
System.out.print(sum[i]);
}
}
【讨论】：
class Sum{
public int number;
public int carry;
Sum(int number, int carry){
this.number = number;
this.carry = carry;
}
}
public String addBinary(String a, String b) {
int lengthOfA = a.length();
int lengthOfB = b.length();
if(lengthOfA > lengthOfB){
for(int i=0; i<(lengthOfA - lengthOfB); i++){
b="0"+b;
}
}
else{
for(int i=0; i<(lengthOfB - lengthOfA); i++){
a="0"+a;
}
}
String result = "";
Sum s = new Sum(0,0);
for(int i=a.length()-1; i>=0; i--){
s = addNumber(Character.getNumericValue(a.charAt(i)), Character.getNumericValue(b.charAt(i)), s.carry);
result = result + Integer.toString(s.number);
}
if(s.carry == 1) { result += s.carry ;}
return new StringBuilder(result).reverse().toString();
}
Sum addNumber(int number1, int number2, int carry){
Sum sum = new Sum(0,0);
sum.number = number1 ^ number2 ^ carry;
sum.carry = (number1 & number2) | (number2 & carry) | (number1 & carry);
return sum;
}
【讨论】：
import java.util.*;
public class BitAddition {
/**
* @param args
*/
public static void main(String[] args) {
// TODO Auto-generated method stub
Scanner sc = new Scanner(System.in);
int len = sc.nextInt();
int[] arr1 = new int[len];
int[] arr2 = new int[len];
int[] sum = new int[len+1];
Arrays.fill(sum, 0);
for(int i=0;i<len;i++){
arr1[i] =sc.nextInt();
}
for(int i=0;i<len;i++){
arr2[i] =sc.nextInt();
}
for(int i=len-1;i>=0;i--){
if(sum[i+1] == 0){
if(arr1[i]!=arr2[i]){
sum[i+1] = 1;
}
else if(arr1[i] ==1 && arr2[i] == 1){
sum[i+1] =0 ;
sum[i] = 1;
}
}
else{
if((arr1[i]!=arr2[i])){
sum[i+1] = 0;
sum[i] = 1;
}
else if(arr1[i] == 1){
sum[i+1] = 1;
sum[i] = 1;
}
}
}
for(int i=0;i<=len;i++){
System.out.print(sum[i]);
}
}
}
【讨论】：
其中一种简单的方法是：
//java代码为as
static String binaryAdd(String a, String b){
int len = 0;
int size = 0;
char[] c1 = a.toCharArray();
char[] c2 = b.toCharArray();
char[] max;
if(c1.length > c2.length){
len = c2.length;
size = c1.length;
max = c1;
}
else
{
len = c1.length;
size = c2.length;
max = c2;
}
StringBuilder sb = new StringBuilder();
int carry = 0;
int p = c1.length - 1;
int q = c2.length - 1;
for(int i=len-1; i>=0; i--){
if(c1[p] == '0' && c2[q] == '0'){
if(carry == 0){
sb.append(0);
carry = 0;
}
else{
sb.append(1);
carry = 0;
}
}
if((c1[p] == '0' && c2[q] == '1') || (c1[p] == '1' && c2[q] == '0')){
if(carry == 0){
sb.append(1);
carry = 0;
}
else{
sb.append(0);
carry = 1;
}
}
if((c1[p] == '1' && c2[q] == '1')){
if(carry == 0){
sb.append(0);
carry = 1;
}
else{
sb.append(1);
carry = 1;
}
}
p--;
q--;
}
for(int j = size-len-1; j>=0; j--){
if(max[j] == '0'){
if(carry == 0){
sb.append(0);
carry = 0;
}
else{
sb.append(1);
carry = 0;
}
}
if(max[j] == '1'){
if(carry == 0){
sb.append(1);
carry = 0;
}
else{
sb.append(0);
carry = 1;
}
}
}
if(carry == 1)
sb.append(1);
return sb.reverse().toString();
}
【讨论】：
import java.io.;
import java.util.;
public class adtbin {
static Scanner sc=new Scanner(System.in);
public void fun(int n1) {
int i=0;
int sum[]=new int[20];
while(n1>0) {
sum[i]=n1%2; n1=n1/2; i++;
}
for(int a=i-1;a>=0;a--) {
System.out.print(sum[a]);
}
}
public static void main() {
int m,n,add;
adtbin ob=new adtbin();
System.out.println("enter the value of m and n");
m=sc.nextInt();
n=sc.nextInt();
add=m+n;
ob.fun(add);
}
}
【讨论】：
你可以自己写一个。
long a =100011111111L;
long b =1000001111L;
int carry = 0 ;
long result = 0;
long multiplicity = 1;
while(a!=0 || b!=0 || carry ==1){
if(a%10==1){
if(b%10==1){
result+= (carry*multiplicity);
carry = 1;
}else if(carry == 1){
carry = 1;
}else{
result += multiplicity;
}
}else if (b%10 == 1){
if(carry == 1){
carry = 1;
}else {
result += multiplicity;
}
}else {
result += (carry*multiplicity);
carry = 0;
}
a/=10;
b/=10;
multiplicity *= 10;
}
System.out.print(result);
它只通过数字工作，不需要字符串，不需要 SubString 和 ...
【讨论】：
package Assignment19thDec;
import java.util.Scanner;
public class addTwoBinaryNumbers {
private static Scanner sc;
public static void main(String[] args) {
sc = new Scanner(System.in);
System.out.println("Enter 1st Binary Number");
int number1=sc.nextInt();
int reminder1=0;
int number2=sc.nextInt();
int reminder2=0;
int carry=0;
double sumResult=0 ;int add = 0
;
int n;
int power=0;
while (number1>0 || number2>0) {
/*System.out.println(number1 + " " +number2);*/
reminder1=number1%10;
number1=number1/10;
reminder2=number2%10;
number2=number2/10;
/*System.out.println(reminder1 +" "+ reminder2);*/
if(reminder1>1 || reminder2>1 ) {
System.out.println("not a binary number");
System.exit(0);
}
n=reminder1+reminder2+carry;
switch(n) {
case 0:
add=0; carry=0;
break;
case 1: add=1; carry=0;
break;
case 2: add=0; carry=1;
break;
case 3: add=1;carry=1;
break;
default: System.out.println("not a binary number ");
}
sumResult=add*(Math.pow(10, power))+sumResult;
power++;
}
sumResult=carry*(Math.pow(10, power))+sumResult;
System.out.println("\n"+(int)sumResult);
}
}
【讨论】：
Try this, tested with binary and decimal and its self explanatory
public String add(String s1, String s2, int radix){
int s1Length = s1.length();
int s2Length = s2.length();
int reminder = 0;
int carry = 0;
StringBuilder result = new StringBuilder();
int i = s1Length -1;
int j = s2Length -1;
while (i >=0 && j>=0) {
int operand1 = Integer.valueOf(s1.charAt(i)+"");
int operand2 = Integer.valueOf(s2.charAt(j)+"");
reminder = (operand1+operand2+carry) % radix;
carry = (operand1+operand2+carry) / radix;
result.append(reminder);
i--;j--;
}
while(i>=0){
int operand1 = Integer.valueOf(s1.charAt(i)+"");
reminder = (operand1+carry) % radix;
carry = (operand1+carry) / radix;
result.append(reminder);
i--;
}
while(j>=0){
int operand1 = Integer.valueOf(s2.charAt(j)+"");
reminder = (operand1+carry) % radix;
carry = (operand1+carry) / radix;
result.append(reminder);
j--;
}
return result.reverse().toString();
}
}
【讨论】：
import java.util.Scanner;
{
public static void main(String[] args)
{
String b1,b2;
Scanner sc= new Scanner(System.in);
System.out.println("Enter 1st binary no. : ") ;
b1=sc.next();
System.out.println("Enter 2nd binary no. : ") ;
b2=sc.next();
int num1=Integer.parseInt(b1,2);
int num2=Integer.parseInt(b2,2);
int sum=num1+num2;
System.out.println("Additon is : "+Integer.toBinaryString(sum));
}
}
【讨论】：