blocks|key|1093680|text|Integer.toBinaryString(int+i)|type|code-block|depth|inlineStyleRanges|entityRanges|data|syntax|javascript|1093681|unstyled|entityMap^0|0^^$0|@$1|2|3|4|5|6|7|G|8|@]|9|@]|A|$B|C]]|$1|D|3|-4|5|E|7|H|8|@]|9|@]|A|$]]]|F|$]]

<pre><code>Integer.toBinaryString(int i)
</code></pre>

blocks|key|1308862|text|还有java.lang.Integer.toString(int+i,+int+base)方法，如果您的代码有一天可以处理2(二进制)之外的基数，它会更合适。请记住，此方法只给出了整数i的无符号表示，如果它是负数，它将在前面添加一个负号。它不会使用二的补码。|type|unstyled|depth|inlineStyleRanges|entityRanges|offset|length|data|1308863|entityMap|0|LINK|mutability|MUTABLE|url|https://docs.oracle.com/javase/7/docs/api/java/lang/Integer.html#toString(int,%2520int)^0|2|17|0|0^^$0|@$1|2|3|4|5|6|7|L|8|@]|9|@$A|M|B|N|1|O]]|C|$]]|$1|D|3|-4|5|6|7|P|8|@]|9|@]|C|$]]]|E|$F|$5|G|H|I|C|$J|K]]]]

There is also the <a href="https://docs.oracle.com/javase/7/docs/api/java/lang/Integer.html#toString(int,%20int)" rel="nofollow noreferrer">java.lang.Integer.toString(int i, int base)</a> method, which would be more appropriate if your code might one day handle bases other than 2 (binary). Keep in mind that this method only gives you an unsigned representation of the integer i, and if it is negative, it will tack on a negative sign at the front. It won't use two's complement.

blocks|key|1093817|text|public+static+string+intToBinary(int+n)
{
++++String+s+=+"";
++++while+(n+>+0)
++++{
++++++++s+=++(+(n+%25+2+)+==+0+?+"0"+:+"1")+%2Bs;
++++++++n+=+n+/+2;
++++}
++++return+s;
}|type|code-block|depth|inlineStyleRanges|entityRanges|data|syntax|javascript|1093818|unstyled|entityMap^0|0^^$0|@$1|2|3|4|5|6|7|G|8|@]|9|@]|A|$B|C]]|$1|D|3|-4|5|E|7|H|8|@]|9|@]|A|$]]]|F|$]]

<pre><code>public static string intToBinary(int n)
{
 String s = &quot;&quot;;
 while (n &gt; 0)
 {
 s = ( (n % 2 ) == 0 ? &quot;0&quot; : &quot;1&quot;) +s;
 n = n / 2;
 }
 return s;
}
</code></pre>

blocks|key|1093713|text|public+class+Main++{

+++public+static+String+toBinary(int+n,+int+l+)+throws+Exception+{
+++++++double+pow+=++Math.pow(2,+l);
+++++++StringBuilder+binary+=+new+StringBuilder();
++++++++if+(+pow+<+n+)+{
++++++++++++throw+new+Exception("The+length+must+be+big+from+number+");
++++++++}
+++++++int+shift+=+l-+1;
+++++++for+(;+shift+>=+0+;+shift--)+{
+++++++++++int+bit+=+(n+>>+shift)+&+1;
+++++++++++if+(bit+==+1)+{
+++++++++++++++binary.append("1");
+++++++++++}+else+{
+++++++++++++++binary.append("0");
+++++++++++}
+++++++}
+++++++return+binary.toString();
+++}

++++public+static+void+main(String[]+args)+throws+Exception+{
++++++++System.out.println("+binary+=+"+%2B+toBinary(7,+4));
++++++++System.out.println("+binary+=+"+%2B+Integer.toString(7,2));
++++}
}|type|code-block|depth|inlineStyleRanges|entityRanges|data|syntax|javascript|1093714|unstyled|entityMap^0|0^^$0|@$1|2|3|4|5|6|7|G|8|@]|9|@]|A|$B|C]]|$1|D|3|-4|5|E|7|H|8|@]|9|@]|A|$]]]|F|$]]

<pre><code>public class Main {

 public static String toBinary(int n, int l ) throws Exception {
 double pow = Math.pow(2, l);
 StringBuilder binary = new StringBuilder();
 if ( pow &lt; n ) {
 throw new Exception("The length must be big from number ");
 }
 int shift = l- 1;
 for (; shift &gt;= 0 ; shift--) {
 int bit = (n &gt;&gt; shift) &amp; 1;
 if (bit == 1) {
 binary.append("1");
 } else {
 binary.append("0");
 }
 }
 return binary.toString();
 }

 public static void main(String[] args) throws Exception {
 System.out.println(" binary = " + toBinary(7, 4));
 System.out.println(" binary = " + Integer.toString(7,2));
 }
}
</code></pre>

blocks|key|1308888|text|这是我几分钟前写的东西，只是随便看看。希望它能帮上忙！|type|unstyled|depth|inlineStyleRanges|entityRanges|data|1308889|public+class+Main+{

public+static+void+main(String[]+args)+{

++++ArrayList<Integer>+powers+=+new+ArrayList<Integer>();
++++ArrayList<Integer>+binaryStore+=+new+ArrayList<Integer>();

++++powers.add(128);
++++powers.add(64);
++++powers.add(32);
++++powers.add(16);
++++powers.add(8);
++++powers.add(4);
++++powers.add(2);
++++powers.add(1);

++++Scanner+sc+=+new+Scanner(System.in);
++++System.out.println("Welcome+to+Paden9000+binary+converter.+Please+enter+an+integer+you+wish+to+convert:+");
++++int+input+=+sc.nextInt();
++++int+printableInput+=+input;

++++for+(int+i+:+powers)+{
++++++++if+(input+<+i)+{
++++++++++++binaryStore.add(0);+++++
++++++++}+else+{
++++++++++++input+=+input+-+i;
++++++++++++binaryStore.add(1);+++++++++++++
++++++++}+++++++++++
++++}

++++String+newString=+binaryStore.toString();
++++String+finalOutput+=+newString.replace("[",+"")
++++++++++++.replace("+",+"")
++++++++++++.replace("]",+"")
++++++++++++.replace(",",+"");

++++System.out.println("Integer+value:+"+%2B+printableInput+%2B+"\nBinary+value:+"+%2B+finalOutput);
++++sc.close();
}+++|code-block|syntax|javascript|1308890|}|1308891|entityMap^0|0|0|0^^$0|@$1|2|3|4|5|6|7|K|8|@]|9|@]|A|$]]|$1|B|3|C|5|D|7|L|8|@]|9|@]|A|$E|F]]|$1|G|3|H|5|6|7|M|8|@]|9|@]|A|$]]|$1|I|3|-4|5|6|7|N|8|@]|9|@]|A|$]]]|J|$]]

This is something I wrote a few minutes ago just messing around. Hope it helps!

<pre><code>public class Main {

public static void main(String[] args) {

 ArrayList&lt;Integer&gt; powers = new ArrayList&lt;Integer&gt;();
 ArrayList&lt;Integer&gt; binaryStore = new ArrayList&lt;Integer&gt;();

 powers.add(128);
 powers.add(64);
 powers.add(32);
 powers.add(16);
 powers.add(8);
 powers.add(4);
 powers.add(2);
 powers.add(1);

 Scanner sc = new Scanner(System.in);
 System.out.println("Welcome to Paden9000 binary converter. Please enter an integer you wish to convert: ");
 int input = sc.nextInt();
 int printableInput = input;

 for (int i : powers) {
 if (input &lt; i) {
 binaryStore.add(0); 
 } else {
 input = input - i;
 binaryStore.add(1); 
 } 
 }

 String newString= binaryStore.toString();
 String finalOutput = newString.replace("[", "")
 .replace(" ", "")
 .replace("]", "")
 .replace(",", "");

 System.out.println("Integer value: " + printableInput + "\nBinary value: " + finalOutput);
 sc.close();
} 
</code></pre>

}

blocks|key|1308920|text|使用内置函数：|type|unstyled|depth|inlineStyleRanges|entityRanges|data|1308921|String+binaryNum+=+Integer.toBinaryString(int+num);|code-block|syntax|javascript|1308922|如果您不想使用内置函数将int转换为二进制，那么您也可以这样做：|1308923|import+java.util.*;
public+class+IntToBinary+{
++++public+static+void+main(String[]+args)+{
++++++++Scanner+d+=+new+Scanner(System.in);
++++++++int+n;
++++++++n+=+d.nextInt();
++++++++StringBuilder+sb+=+new+StringBuilder();
++++++++while(n+>+0){
++++++++int+r+=+n%252;
++++++++sb.append(r);
++++++++n+=+n/2;
++++++++}
++++++++System.out.println(sb.reverse());++++++++
++++}
}|1308924|entityMap^0|0|0|0|0^^$0|@$1|2|3|4|5|6|7|M|8|@]|9|@]|A|$]]|$1|B|3|C|5|D|7|N|8|@]|9|@]|A|$E|F]]|$1|G|3|H|5|6|7|O|8|@]|9|@]|A|$]]|$1|I|3|J|5|D|7|P|8|@]|9|@]|A|$E|F]]|$1|K|3|-4|5|6|7|Q|8|@]|9|@]|A|$]]]|L|$]]

Using built-in function:

<pre><code>String binaryNum = Integer.toBinaryString(int num);
</code></pre>

If you don't want to use the built-in function for converting int to binary then you can also do this:

<pre><code>import java.util.*;
public class IntToBinary {
 public static void main(String[] args) {
 Scanner d = new Scanner(System.in);
 int n;
 n = d.nextInt();
 StringBuilder sb = new StringBuilder();
 while(n &gt; 0){
 int r = n%2;
 sb.append(r);
 n = n/2;
 }
 System.out.println(sb.reverse()); 
 }
}
</code></pre>

blocks|key|1308966|text|最简单的方法是检查数字是否为奇数。如果是，根据定义，它最右边的二进制数将是"1“(2%5E0)。在我们确定这一点之后，我们将数字向右移位，并使用递归检查相同的值。|type|unstyled|depth|inlineStyleRanges|entityRanges|data|1308967|@Test
public+void+shouldPrintBinary()+{
++++StringBuilder+sb+=+new+StringBuilder();
++++convert(1234,+sb);
}

private+void+convert(int+n,+StringBuilder+sb)+{

++++if+(n+>+0)+{
++++++++sb.append(n+%25+2);
++++++++convert(n+>>+1,+sb);
++++}+else+{
++++++++System.out.println(sb.reverse().toString());
++++}
}|code-block|syntax|javascript|1308968|entityMap^0|0|0^^$0|@$1|2|3|4|5|6|7|I|8|@]|9|@]|A|$]]|$1|B|3|C|5|D|7|J|8|@]|9|@]|A|$E|F]]|$1|G|3|-4|5|6|7|K|8|@]|9|@]|A|$]]]|H|$]]

The simplest approach is to check whether or not the number is odd. If it is, by definition, its right-most binary number will be "1" (2^0). After we've determined this, we bit shift the number to the right and check the same value using recursion.

<pre><code>@Test
public void shouldPrintBinary() {
 StringBuilder sb = new StringBuilder();
 convert(1234, sb);
}

private void convert(int n, StringBuilder sb) {

 if (n &gt; 0) {
 sb.append(n % 2);
 convert(n &gt;&gt; 1, sb);
 } else {
 System.out.println(sb.reverse().toString());
 }
}
</code></pre>

blocks|key|1093932|text|这里是我的方法，它是有点令人信服的字节数固定|type|unstyled|depth|inlineStyleRanges|entityRanges|data|1093933|private+void+printByte(int+value)+{
String+currentBinary+=+Integer.toBinaryString(256+%2B+value);
System.out.println(currentBinary.substring(currentBinary.length()+-+8));
}

public+int+binaryToInteger(String+binary)+{
char[]+numbers+=+binary.toCharArray();
int+result+=+0;
for(int+i=numbers.length+-+1;+i>=0;+i--)
++if(numbers[i]=='1')
++++result+%2B=+Math.pow(2,+(numbers.length-i+-+1));
return+result;
}|code-block|syntax|javascript|1093934|entityMap^0|0|0^^$0|@$1|2|3|4|5|6|7|I|8|@]|9|@]|A|$]]|$1|B|3|C|5|D|7|J|8|@]|9|@]|A|$E|F]]|$1|G|3|-4|5|6|7|K|8|@]|9|@]|A|$]]]|H|$]]

here is my methods, it is a little bit convince that number of bytes fixed

<pre><code>private void printByte(int value) {
String currentBinary = Integer.toBinaryString(256 + value);
System.out.println(currentBinary.substring(currentBinary.length() - 8));
}

public int binaryToInteger(String binary) {
char[] numbers = binary.toCharArray();
int result = 0;
for(int i=numbers.length - 1; i&gt;=0; i--)
 if(numbers[i]=='1')
 result += Math.pow(2, (numbers.length-i - 1));
return result;
}
</code></pre>

blocks|key|1094042|text|使用位移位会更快一点...|type|unstyled|depth|inlineStyleRanges|entityRanges|data|1094043|public+static+String+convertDecimalToBinary(int+N)+{

++++StringBuilder+binary+=+new+StringBuilder(32);

++++while+(N+>+0+)+{
++++++++binary.append(+N+%25+2+);
++++++++N+>>=+1;
+++++}

++++return+binary.reverse().toString();

}|code-block|syntax|javascript|1094044|entityMap^0|0|0^^$0|@$1|2|3|4|5|6|7|I|8|@]|9|@]|A|$]]|$1|B|3|C|5|D|7|J|8|@]|9|@]|A|$E|F]]|$1|G|3|-4|5|6|7|K|8|@]|9|@]|A|$]]]|H|$]]

Using bit shift is a little quicker... 

<pre><code>public static String convertDecimalToBinary(int N) {

 StringBuilder binary = new StringBuilder(32);

 while (N &gt; 0 ) {
 binary.append( N % 2 );
 N &gt;&gt;= 1;
 }

 return binary.reverse().toString();

}
</code></pre>

blocks|key|1309063|text|这可以用伪代码表示为：|type|unstyled|depth|inlineStyleRanges|entityRanges|data|1309064|while(n+>+0):
++++remainder+=+n%252;
++++n+=+n/2;
++++Insert+remainder+to+front+of+a+list+or+push+onto+a+stack

Print+list+or+stack|code-block|syntax|javascript|1309065|entityMap^0|0|0^^$0|@$1|2|3|4|5|6|7|I|8|@]|9|@]|A|$]]|$1|B|3|C|5|D|7|J|8|@]|9|@]|A|$E|F]]|$1|G|3|-4|5|6|7|K|8|@]|9|@]|A|$]]]|H|$]]

This can be expressed in pseudocode as:

<pre><code>while(n &gt; 0):
 remainder = n%2;
 n = n/2;
 Insert remainder to front of a list or push onto a stack

Print list or stack
</code></pre>

blocks|key|1309166|text|你真的应该使用Integer.toBinaryString()+(如上所示)，但是如果出于某种原因你想要自己的：|type|unstyled|depth|inlineStyleRanges|entityRanges|offset|length|data|1309167|//+Like+Integer.toBinaryString,+but+always+returns+32+chars
public+static+String+asBitString(int+value)+{
++final+char[]+buf+=+new+char[32];
++for+(int+i+=+31;+i+>=+0;+i--)+{
++++buf[31+-+i]+=+((1+<<+i)+&+value)+==+0+?+'0'+:+'1';
++}
++return+new+String(buf);
}|code-block|syntax|javascript|1309168|entityMap|0|LINK|mutability|MUTABLE|url|https://docs.oracle.com/javase/8/docs/api/java/lang/Integer.html#toBinaryString-int-^0|7|O|0|0|0^^$0|@$1|2|3|4|5|6|7|Q|8|@]|9|@$A|R|B|S|1|T]]|C|$]]|$1|D|3|E|5|F|7|U|8|@]|9|@]|C|$G|H]]|$1|I|3|-4|5|6|7|V|8|@]|9|@]|C|$]]]|J|$K|$5|L|M|N|C|$O|P]]]]

You should really use <a href="https://docs.oracle.com/javase/8/docs/api/java/lang/Integer.html#toBinaryString-int-" rel="nofollow noreferrer">Integer.toBinaryString()</a> (as shown above), but if for some reason you want your own:

<pre><code>// Like Integer.toBinaryString, but always returns 32 chars
public static String asBitString(int value) {
 final char[] buf = new char[32];
 for (int i = 31; i &gt;= 0; i--) {
 buf[31 - i] = ((1 &lt;&lt; i) &amp; value) == 0 ? '0' : '1';
 }
 return new String(buf);
}
</code></pre>

blocks|key|1309383|text|如果int值是15，您可以将其转换为二进制，如下所示。|type|unstyled|depth|inlineStyleRanges|entityRanges|data|1309384|int+x+=+15;

Integer.toBinaryString(x);|code-block|syntax|javascript|1309385|如果您有二进制值，您可以将其转换为int值，如下所示。|1309386|String+binaryValue+=+"1010";

Integer.parseInt(binaryValue,+2);|1309387|entityMap^0|0|0|0|0^^$0|@$1|2|3|4|5|6|7|M|8|@]|9|@]|A|$]]|$1|B|3|C|5|D|7|N|8|@]|9|@]|A|$E|F]]|$1|G|3|H|5|6|7|O|8|@]|9|@]|A|$]]|$1|I|3|J|5|D|7|P|8|@]|9|@]|A|$E|F]]|$1|K|3|-4|5|6|7|Q|8|@]|9|@]|A|$]]]|L|$]]

if the int value is 15, you can convert it to a binary as follows.
<pre><code>int x = 15;

Integer.toBinaryString(x);
</code></pre>
if you have the binary value, you can convert it into int value as follows.
<pre><code>String binaryValue = &quot;1010&quot;;

Integer.parseInt(binaryValue, 2);
</code></pre>

blocks|key|1309024|text|这应该很简单，如下所示：|type|unstyled|depth|inlineStyleRanges|entityRanges|data|1309025|public+static+String+toBinary(int+number){
++++StringBuilder+sb+=+new+StringBuilder();

++++if(number+==+0)
++++++++return+"0";
++++while(number>=1){
++++++++sb.append(number%252);
++++++++number+=+number+/+2;
++++}

++++return+sb.reverse().toString();

}|code-block|syntax|javascript|1309026|entityMap^0|0|0^^$0|@$1|2|3|4|5|6|7|I|8|@]|9|@]|A|$]]|$1|B|3|C|5|D|7|J|8|@]|9|@]|A|$E|F]]|$1|G|3|-4|5|6|7|K|8|@]|9|@]|A|$]]]|H|$]]

This should be quite simple with something like this :

<pre><code>public static String toBinary(int number){
 StringBuilder sb = new StringBuilder();

 if(number == 0)
 return "0";
 while(number&gt;=1){
 sb.append(number%2);
 number = number / 2;
 }

 return sb.reverse().toString();

}
</code></pre>

blocks|key|1309137|text|您还可以使用while循环将int转换为二进制。像这样，|type|unstyled|depth|inlineStyleRanges|entityRanges|data|1309138|import+java.util.Scanner;

public+class+IntegerToBinary
{
+++public+static+void+main(String[]+args)
+++{
++++++int+num;
++++++String+str+=+"";
++++++Scanner+sc+=+new+Scanner(System.in);
++++++System.out.print("Please+enter+the+a+number+:+");
++++++num+=+sc.nextInt();
++++++while(num+>+0)
++++++{
+++++++++int+y+=+num+%25+2;
+++++++++str+=+y+%2B+str;
+++++++++num+=+num+/+2;
++++++}
++++++System.out.println("The+binary+conversion+is+:+"+%2B+str);
++++++sc.close();
+++}
}|code-block|syntax|javascript|1309139|源代码和参考资料-+convert+int+to+binary+in+java示例。|offset|length|1309140|entityMap|0|LINK|mutability|MUTABLE|url|https://www.flowerbrackets.com/java-program-to-convert-integer-value-into-binary/^0|0|0|A|T|0|0^^$0|@$1|2|3|4|5|6|7|S|8|@]|9|@]|A|$]]|$1|B|3|C|5|D|7|T|8|@]|9|@]|A|$E|F]]|$1|G|3|H|5|6|7|U|8|@]|9|@$I|V|J|W|1|X]]|A|$]]|$1|K|3|-4|5|6|7|Y|8|@]|9|@]|A|$]]]|L|$M|$5|N|O|P|A|$Q|R]]]]

You can use while loop as well to convert an int to binary. Like this,
<pre><code>import java.util.Scanner;

public class IntegerToBinary
{
 public static void main(String[] args)
 {
 int num;
 String str = &quot;&quot;;
 Scanner sc = new Scanner(System.in);
 System.out.print(&quot;Please enter the a number : &quot;);
 num = sc.nextInt();
 while(num &gt; 0)
 {
 int y = num % 2;
 str = y + str;
 num = num / 2;
 }
 System.out.println(&quot;The binary conversion is : &quot; + str);
 sc.close();
 }
}
</code></pre>
Source and reference - <a href="https://www.flowerbrackets.com/java-program-to-convert-integer-value-into-binary/" rel="nofollow noreferrer">convert int to binary in java</a> example.

blocks|key|1094053|text|public+class+BinaryConverter+{

++++public+static+String+binaryConverter(int+number)+{
++++++++String+binary+=+"";
++++++++if+(number+==+1){
++++++++++++binary+=+"1";
++++++++++++return+binary;
++++++++}
++++++++if+(number+==+0){
++++++++++++binary+=+"0";
++++++++++++return+binary;
++++++++}
++++++++if+(number+>+1)+{
++++++++++++String+i+=+Integer.toString(number+%25+2);

++++++++++++binary+=+binary+%2B+i;
++++++++++++binaryConverter(number/2);
++++++++}
++++++++return+binary;
++++}
}|type|code-block|depth|inlineStyleRanges|entityRanges|data|syntax|javascript|1094054|unstyled|entityMap^0|0^^$0|@$1|2|3|4|5|6|7|G|8|@]|9|@]|A|$B|C]]|$1|D|3|-4|5|E|7|H|8|@]|9|@]|A|$]]]|F|$]]

<pre><code>public class BinaryConverter {

 public static String binaryConverter(int number) {
 String binary = &quot;&quot;;
 if (number == 1){
 binary = &quot;1&quot;;
 return binary;
 }
 if (number == 0){
 binary = &quot;0&quot;;
 return binary;
 }
 if (number &gt; 1) {
 String i = Integer.toString(number % 2);

 binary = binary + i;
 binaryConverter(number/2);
 }
 return binary;
 }
}
</code></pre>

blocks|key|1094119|text|为了精确到8位，我对@sandeep-saini的回答做了一点补充：|type|unstyled|depth|inlineStyleRanges|entityRanges|data|1094120|public+static+String+intToBinary(int+number){
++++++++StringBuilder+sb+=+new+StringBuilder();

++++++++if(number+==+0)
++++++++++++return+"0";
++++++++while(number>=1){
++++++++++++sb.append(number%252);
++++++++++++number+=+number+/+2;
++++++++}
++++++++while+(sb.length()+<+8){
++++++++++++sb.append("0");
++++++++}

++++++++return+sb.reverse().toString();

++++}|code-block|syntax|javascript|1094121|现在，对于1的输入，您将得到00000001的输出|offset|length|style|CODE|1094122|entityMap^0|0|0|5|1|E|8|0^^$0|@$1|2|3|4|5|6|7|O|8|@]|9|@]|A|$]]|$1|B|3|C|5|D|7|P|8|@]|9|@]|A|$E|F]]|$1|G|3|H|5|6|7|Q|8|@$I|R|J|S|K|L]|$I|T|J|U|K|L]]|9|@]|A|$]]|$1|M|3|-4|5|6|7|V|8|@]|9|@]|A|$]]]|N|$]]

In order to make it exactly 8 bit, I made a slight addition to @sandeep-saini 's answer:
<pre><code>public static String intToBinary(int number){
 StringBuilder sb = new StringBuilder();

 if(number == 0)
 return &quot;0&quot;;
 while(number&gt;=1){
 sb.append(number%2);
 number = number / 2;
 }
 while (sb.length() &lt; 8){
 sb.append(&quot;0&quot;);
 }

 return sb.reverse().toString();

 }
</code></pre>
So now for an input of <code>1</code> you get an output of <code>00000001</code>

blocks|key|1309327|text|public+static+String+intToBinaryString(int+n)+{
++++StringBuilder+sb+=+new+StringBuilder();
++++for+(int+i+=+0;+i+<+32+&&+n+!=+0;+i%2B%2B)+{
++++++++sb.append((n&1)+==+1+?+"1"+:+"0");
++++++++n+>>=+1;
++++}
++++return+sb.reverse().toString();
}|type|code-block|depth|inlineStyleRanges|entityRanges|data|syntax|javascript|1309328|我们不能使用n%252来检查第一位，因为它不适用于负整数。我们应该使用n&1。|unstyled|offset|length|style|CODE|1309329|entityMap^0|0|6|3|X|3|0^^$0|@$1|2|3|4|5|6|7|M|8|@]|9|@]|A|$B|C]]|$1|D|3|E|5|F|7|N|8|@$G|O|H|P|I|J]|$G|Q|H|R|I|J]]|9|@]|A|$]]|$1|K|3|-4|5|F|7|S|8|@]|9|@]|A|$]]]|L|$]]

<pre><code>public static String intToBinaryString(int n) {
 StringBuilder sb = new StringBuilder();
 for (int i = 0; i &lt; 32 &amp;&amp; n != 0; i++) {
 sb.append((n&amp;1) == 1 ? &quot;1&quot; : &quot;0&quot;);
 n &gt;&gt;= 1;
 }
 return sb.reverse().toString();
}
</code></pre>
We cannot use <code>n%2</code> to check the first bit, because it's not right for negtive integer. We should use <code>n&amp;1</code>.

blocks|key|1309428|text|我的2cents：|type|unstyled|depth|inlineStyleRanges|entityRanges|data|1309429|public+class+Integer2Binary+{
++++public+static+void+main(String[]+args)+{
++++++++final+int+integer12+=+12;
++++++++System.out.println(integer12+%2B+"+->+"+%2B+integer2Binary(integer12));
++++++++//+12+->+00000000000000000000000000001100
++++}

++++private+static+String+integer2Binary(int+n)+{
++++++++return+new+StringBuilder(Integer.toBinaryString(n))
++++++++++++.insert(0,+"0".repeat(Integer.numberOfLeadingZeros(n)))
++++++++++++.toString();
++++}
}|code-block|syntax|javascript|1309430|entityMap^0|0|0^^$0|@$1|2|3|4|5|6|7|I|8|@]|9|@]|A|$]]|$1|B|3|C|5|D|7|J|8|@]|9|@]|A|$E|F]]|$1|G|3|-4|5|6|7|K|8|@]|9|@]|A|$]]]|H|$]]

My 2cents:
<pre><code>public class Integer2Binary {
 public static void main(String[] args) {
 final int integer12 = 12;
 System.out.println(integer12 + &quot; -&gt; &quot; + integer2Binary(integer12));
 // 12 -&gt; 00000000000000000000000000001100
 }

 private static String integer2Binary(int n) {
 return new StringBuilder(Integer.toBinaryString(n))
 .insert(0, &quot;0&quot;.repeat(Integer.numberOfLeadingZeros(n)))
 .toString();
 }
}
</code></pre>

What would be the best way (ideally, simplest) to convert an int to a binary string representation in Java?

For example, say the int is 156. The binary string representation of this would be "10011100".

Converting an int to a binary string representation in Java?

Java

在Java中，将int转换为二进制字符串表示的最佳方式(理想情况下，最简单)是什么？例如，假设int是156。它的二进制字符串表示应该是"10011100“。

问在Java中将int转换为二进制字符串表示？
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回答 18

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回答 18

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