blocks|key|2485821|text|试试这个：|type|unstyled|depth|inlineStyleRanges|entityRanges|data|2485822|//Edit+your+min+and+max,+e.g.+Integer.MAX_VALUE+and+Integer.MIN_VALUE
int+min+=+0;
int+max+=+10;

for(int+i+=+min;+i+<+max;+i%2B%2B){
++++//Make+16+bit+long+binary+Strings
++++String+s+=+String.format("%2516s",+Integer.toBinaryString(i)).replace('+',+'0');

++++//Make+4+bits+long+chunks
++++List<String>+chunks+=+new+ArrayList<>();
++++Matcher+matcher+=+Pattern.compile(".{0,4}").matcher(s);
++++while+(matcher.find())+{
++++++++chunks.add(s.substring(matcher.start(),+matcher.end()));
++++}

++++StringBuilder+b+=+new+StringBuilder();
++++for+(String+c+:+chunks)+{
++++++++//Here+you+can+count+the+1+and+0+of+the+current+chunk+with+c.charAt(index)
++++++++b.append(c);
++++++++b.append("+");
++++}
++++System.out.println(b.toString());
}|code-block|syntax|javascript|2485823|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|$]]

Try this:

<pre><code>//Edit your min and max, e.g. Integer.MAX_VALUE and Integer.MIN_VALUE
int min = 0;
int max = 10;

for(int i = min; i &lt; max; i++){
 //Make 16 bit long binary Strings
 String s = String.format("%16s", Integer.toBinaryString(i)).replace(' ', '0');

 //Make 4 bits long chunks
 List&lt;String&gt; chunks = new ArrayList&lt;&gt;();
 Matcher matcher = Pattern.compile(".{0,4}").matcher(s);
 while (matcher.find()) {
 chunks.add(s.substring(matcher.start(), matcher.end()));
 }

 StringBuilder b = new StringBuilder();
 for (String c : chunks) {
 //Here you can count the 1 and 0 of the current chunk with c.charAt(index)
 b.append(c);
 b.append(" ");
 }
 System.out.println(b.toString());
}
</code></pre>

blocks|key|2433402|text|如果要保留0，只需添加填充：|type|unstyled|depth|inlineStyleRanges|entityRanges|data|2433403|int+end+=+100;+//Change+this
for+(int+i+=+0;+i+<=+end;+i%2B%2B)+{
++++String+bytestring+=+Integer.toBinaryString(i);
++++String+padding+=+"00000000000000000000000000000000";
++++bytestring+=+padding.substring(0,+32+-+bytestring.length())+%2B+bytestring;
++++System.out.println(bytestring);
}|code-block|syntax|javascript|2433404|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|$]]

If you want to preserve the 0s, then just add padding:

<pre><code>int end = 100; //Change this
for (int i = 0; i &lt;= end; i++) {
 String bytestring = Integer.toBinaryString(i);
 String padding = "00000000000000000000000000000000";
 bytestring = padding.substring(0, 32 - bytestring.length()) + bytestring;
 System.out.println(bytestring);
}
</code></pre>

blocks|key|2769356|text|这可以通过递归函数调用来解决：|type|unstyled|depth|inlineStyleRanges|entityRanges|data|2769357|public+class+BinarySequences+{

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

++++++++final+int+numCount+=+4;+
++++++++final+int+oneCount+=+2;+

++++++++checkSubString(numCount,+oneCount,+"");

++++++++for+(String+res+:+results)+{
++++++++++++System.out.println(res);
++++++++}
++++}

++++private+static+List<String>+results+=+new+ArrayList<>();

++++private+static+void+checkSubString(int+numCount,+int+oneCount,+String+prefix)+{
++++++++if+((numCount+>=+oneCount)+&&+(oneCount+>=+0))+{
++++++++++++if+(numCount==1)+{
++++++++++++++++if+(oneCount==1)+{
++++++++++++++++++++results.add(prefix+%2B+"1");
++++++++++++++++}+else+{
++++++++++++++++++++results.add(prefix+%2B+"0");
++++++++++++++++}
++++++++++++}+else+{
++++++++++++++++checkSubString(numCount-1,+oneCount++,+prefix+%2B+"0");
++++++++++++++++checkSubString(numCount-1,+oneCount-1,+prefix+%2B+"1");
++++++++++++}
++++++++}
++++}


}+|code-block|syntax|javascript|2769358|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 solved using recursive function calls:

<pre><code>public class BinarySequences {

 public static void main(String[] args) {

 final int numCount = 4; 
 final int oneCount = 2; 

 checkSubString(numCount, oneCount, "");

 for (String res : results) {
 System.out.println(res);
 }
 }

 private static List&lt;String&gt; results = new ArrayList&lt;&gt;();

 private static void checkSubString(int numCount, int oneCount, String prefix) {
 if ((numCount &gt;= oneCount) &amp;&amp; (oneCount &gt;= 0)) {
 if (numCount==1) {
 if (oneCount==1) {
 results.add(prefix + "1");
 } else {
 results.add(prefix + "0");
 }
 } else {
 checkSubString(numCount-1, oneCount , prefix + "0");
 checkSubString(numCount-1, oneCount-1, prefix + "1");
 }
 }
 }


} 
</code></pre>

blocks|key|2485871|text|需要org.apache.commons.lang.StringUtils，但这使得它很短：|type|unstyled|depth|inlineStyleRanges|offset|length|style|CODE|entityRanges|data|2485872|final+int+digits+=+4;
final+int+onesrequired+=+2;

int+maxindex+=+(int)+Math.pow(2,+digits);

for+(int+i+=+0;+i+<+maxindex;+i%2B%2B)+{
++++String+binaryStr+=+Integer.toBinaryString(i);

++++if+(StringUtils.countMatches(binaryStr,+"1")+==+onesrequired)+{
++++++++System.out.print(String.format("%25"+%2B+digits+%2B+"s",+binaryStr).replace('+',+'0')+%2B+"+");
++++}
}|code-block|syntax|javascript|2485873|entityMap^0|2|Z|0|0^^$0|@$1|2|3|4|5|6|7|M|8|@$9|N|A|O|B|C]]|D|@]|E|$]]|$1|F|3|G|5|H|7|P|8|@]|D|@]|E|$I|J]]|$1|K|3|-4|5|6|7|Q|8|@]|D|@]|E|$]]]|L|$]]

Requires <code>org.apache.commons.lang.StringUtils</code>, but this makes it a short one:

<pre><code>final int digits = 4;
final int onesrequired = 2;

int maxindex = (int) Math.pow(2, digits);

for (int i = 0; i &lt; maxindex; i++) {
 String binaryStr = Integer.toBinaryString(i);

 if (StringUtils.countMatches(binaryStr, "1") == onesrequired) {
 System.out.print(String.format("%" + digits + "s", binaryStr).replace(' ', '0') + " ");
 }
}
</code></pre>

I want to generate every possible binary sequence of numbers, where each sequence in the list is limited to a specific number of 1's and there is padding of zeros to make every list the same length. 

For example, if the sequence is supposed to be 4 numbers long, and have 2 ones, all sequences would be:

<code>1100 1010 1001 0110 0101 0011</code>

and the zeros at the front of the number are preserved.

Generating Binary Sequences

翻译质量差，导致语言生硬或混乱。

没有提供实际的解决方法或示例。

解答不清晰，无法理解或解决问题。

页面排版不美观，阅读体验差。

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我想要生成每一个可能的二进制数字序列，其中列表中的每个序列都被限制为1的特定数，并且有填充的零使每个列表都具有相同的长度。例如，如果序列应该是4个数字，并且有2个数字，那么所有序列都是：1100  1010  1001  0110 0101 0011数字前面的零被保留下来。

问生成二进制序列
EN

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