blocks|key|82626|text|也许你可以使用一个递归算法：|type|unstyled|depth|inlineStyleRanges|entityRanges|data|82627|public+void+printBin(String+soFar,+int+iterations)+{
++++if(iterations+==+0)+{
++++++++System.out.println(soFar);
++++}
++++else+{
++++++++printBin(soFar+%2B+"0",+iterations+-+1);
++++++++printBin(soFar+%2B+"1",+iterations+-+1);
++++}
}|code-block|syntax|javascript|82628|您可以像这样执行以下代码：|82629|printBin("",+4);|82630|这将为您提供所有可能的4位二进制数。|82631|希望这能有所帮助！|82632|entityMap^0|0|0|0|0|0|0^^$0|@$1|2|3|4|5|6|7|Q|8|@]|9|@]|A|$]]|$1|B|3|C|5|D|7|R|8|@]|9|@]|A|$E|F]]|$1|G|3|H|5|6|7|S|8|@]|9|@]|A|$]]|$1|I|3|J|5|D|7|T|8|@]|9|@]|A|$E|F]]|$1|K|3|L|5|6|7|U|8|@]|9|@]|A|$]]|$1|M|3|N|5|6|7|V|8|@]|9|@]|A|$]]|$1|O|3|-4|5|6|7|W|8|@]|9|@]|A|$]]]|P|$]]

Maybe you could use a recursive algorithm:

<pre><code>public void printBin(String soFar, int iterations) {
 if(iterations == 0) {
 System.out.println(soFar);
 }
 else {
 printBin(soFar + "0", iterations - 1);
 printBin(soFar + "1", iterations - 1);
 }
}
</code></pre>

You would execute this like this:

<pre><code>printBin("", 4);
</code></pre>

That would give you all possible binary numbers with 4 digits.

Hope this helped!

blocks|key|5356571|text|对于一个n-bit二进制数，有2%5En个“排列”。您只需要将整数从0循环到(1<<n)-1，并将每个整数转换为二进制。|type|unstyled|depth|inlineStyleRanges|offset|length|style|CODE|entityRanges|data|5356572|entityMap^0|4|1|H|1|10|8|0^^$0|@$1|2|3|4|5|6|7|H|8|@$9|I|A|J|B|C]|$9|K|A|L|B|C]|$9|M|A|N|B|C]]|D|@]|E|$]]|$1|F|3|-4|5|6|7|O|8|@]|D|@]|E|$]]]|G|$]]

For an <code>n</code>-bit binary number, there are 2^<code>n</code> "permutations". You just need to loop over the integers from 0 to <code>(1&lt;&lt;n)-1</code>, and convert each one to binary.

blocks|key|5356648|text|for(int+i=0;+i+<+128;+i%2B%2B){
++System.out.println(Integer.toBinaryString(i));
}|type|code-block|depth|inlineStyleRanges|entityRanges|data|syntax|javascript|5356649|将最大值调整为您想要的最高值。|unstyled|5356650|如果你需要填充0，今天还有另一个问题：Pad+a+binary+String+equal+to+zero+("0")+with+leading+zeros+in+Java|offset|length|5356651|entityMap|0|LINK|mutability|MUTABLE|url|https://stackoverflow.com/questions/8457734/pad-a-binary-string-equal-to-zero-0-with-leading-zeros-in-java^0|0|0|J|1U|0|0^^$0|@$1|2|3|4|5|6|7|S|8|@]|9|@]|A|$B|C]]|$1|D|3|E|5|F|7|T|8|@]|9|@]|A|$]]|$1|G|3|H|5|F|7|U|8|@]|9|@$I|V|J|W|1|X]]|A|$]]|$1|K|3|-4|5|F|7|Y|8|@]|9|@]|A|$]]]|L|$M|$5|N|O|P|A|$Q|R]]]]

<pre><code>for(int i=0; i &lt; 128; i++){
 System.out.println(Integer.toBinaryString(i));
}
</code></pre>

Adjust the max for as high as you'd like to go.

If you need the padded 0's, there was another question on that just today: <a href="https://stackoverflow.com/questions/8457734/pad-a-binary-string-equal-to-zero-0-with-leading-zeros-in-java">Pad a binary String equal to zero (&quot;0&quot;) with leading zeros in Java</a>

blocks|key|1099603|text|它有助于了解有多少种可能性。|type|unstyled|depth|inlineStyleRanges|entityRanges|data|1099604|2%5E4+=+16，对吧？|offset|length|style|CODE|1099605|了解this也会有所帮助。|1099606|我是这样做的：|1099607|/**
+*+BinaryDemo
+*+@author+Michael
+*+@since+12/10/11
+*/
public+class+BinaryDemo+{

++++public+static+void+main(String[]+args)+{
++++++++if+(args.length+>+0)+{
++++++++++++int+n+=+Integer.parseInt(args[0]);
++++++++++++int+m+=+1;
++++++++++++for+(int+i+=+1;+i+<=+n;+%2B%2Bi)+{
++++++++++++++++m+*=+2;
++++++++++++}
++++++++++++System.out.println("#+bits++:+"+%2B+n);
++++++++++++System.out.println("#+values:+"+%2B+m);
++++++++++++String+format+=+"%25"+%2B+n+%2B+"s";
++++++++++++for+(int+i+=+0;+i+<+m;+%2B%2Bi)+{
++++++++++++++++System.out.println(String.format(format,+Integer.toString(i,+2)));
++++++++++++}

++++++++}+else+{
++++++++++++System.out.println("Usage:+BinaryDemo+<n>");
++++++++}
++++}
}|code-block|syntax|javascript|1099608|entityMap|0|LINK|mutability|MUTABLE|url|http://www.exampledepot.com/egs/java.lang/ConvertNum2.html^0|0|0|8|0|2|4|0|0|0|0^^$0|@$1|2|3|4|5|6|7|Y|8|@]|9|@]|A|$]]|$1|B|3|C|5|6|7|Z|8|@$D|10|E|11|F|G]]|9|@]|A|$]]|$1|H|3|I|5|6|7|12|8|@]|9|@$D|13|E|14|1|15]]|A|$]]|$1|J|3|K|5|6|7|16|8|@]|9|@]|A|$]]|$1|L|3|M|5|N|7|17|8|@]|9|@]|A|$O|P]]|$1|Q|3|-4|5|6|7|18|8|@]|9|@]|A|$]]]|R|$S|$5|T|U|V|A|$W|X]]]]

It helps to know how many possibilities there are. 

<code>2^4 = 16</code>, right?

It'll help to know <a href="http://www.exampledepot.com/egs/java.lang/ConvertNum2.html" rel="nofollow">this</a> as well.

Here's how I'd do it:

<pre><code>/**
 * BinaryDemo
 * @author Michael
 * @since 12/10/11
 */
public class BinaryDemo {

 public static void main(String[] args) {
 if (args.length &gt; 0) {
 int n = Integer.parseInt(args[0]);
 int m = 1;
 for (int i = 1; i &lt;= n; ++i) {
 m *= 2;
 }
 System.out.println("# bits : " + n);
 System.out.println("# values: " + m);
 String format = "%" + n + "s";
 for (int i = 0; i &lt; m; ++i) {
 System.out.println(String.format(format, Integer.toString(i, 2)));
 }

 } else {
 System.out.println("Usage: BinaryDemo &lt;n&gt;");
 }
 }
}
</code></pre>

blocks|key|3741313|text|我的解决方案使用回溯算法。只需使用'0‘和'1’打印所有组合即可。这并不比公认的答案更好，但这只是解决problem.As的另一种方法，示例只需从main()|type|unstyled|depth|inlineStyleRanges|entityRanges|data|3741314|ArrayList<Integer>+chosen+=+new+ArrayList<>();+
printAllBinaryDigits(4,+chosen);

private+static+void+printAllBinaryDigits(int+digits,+ArrayList<Integer>chosen)+{
++++++++++++if+(digits+==+0)+{
++++++++++++++++System.out.println(chosen);
++++++++++++}+else+{
++++++++++++++++chosen.add(0);
++++++++++++++++printAllBinaryDigits(digits-1,+chosen);
++++++++++++++++chosen.remove(chosen.size()-1);
++++++++++++++++chosen.add(1);
++++++++++++++++printAllBinaryDigits(digits-1,+chosen);
++++++++++++++++chosen.remove(chosen.size()-1);
++++++++++++}

++++++++}|code-block|syntax|javascript|3741315|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 solutions uses the backtracking algorithm. Just print all combinations using '0' and then '1'.
It's not better than accepted answer but just one more way to solve the problem.As an example just call the function like below from main() 

<pre><code>ArrayList&lt;Integer&gt; chosen = new ArrayList&lt;&gt;(); 
printAllBinaryDigits(4, chosen);

private static void printAllBinaryDigits(int digits, ArrayList&lt;Integer&gt;chosen) {
 if (digits == 0) {
 System.out.println(chosen);
 } else {
 chosen.add(0);
 printAllBinaryDigits(digits-1, chosen);
 chosen.remove(chosen.size()-1);
 chosen.add(1);
 printAllBinaryDigits(digits-1, chosen);
 chosen.remove(chosen.size()-1);
 }

 }
</code></pre>

blocks|key|5356775|text|下面是在java中使用递归作为方法的解决方案。|type|unstyled|depth|inlineStyleRanges|entityRanges|data|5356776|public+class+NumberOfBinaryPatterns+{
++++static+int[]+bitArray+=+new+int[]{0,1};
++++public+static+void+main(String+args[])
++++{+++
++++++++System.out.println("Below+are+the+patterns\n");
++++++++int+numberOfBits=3;+//+It+can+be+any+value+based+on+the+requirement,+i.e.+4,5,+etc.+We+can+get+the+input+for+"numberOfBits"+from+user+on+runtime+to+enhance+this+further
++++++++drawBinaryPattern(numberOfBits,"");
++++}
++++private+static+void+drawBinaryPattern(int+n,String+seed)
++++{
++++++++if(n==0){
++++++++++++System.out.println(seed);
++++++++++++return;
++++++++}+++
++++++++for(int+j=0;j<bitArray.length;j%2B%2B)
++++++++{
++++++++++++String+temp+=+seed%2BbitArray[j];
++++++++++++drawBinaryPattern(n-1,temp);
++++++++}
++++}
}|code-block|syntax|javascript|5356777|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|$]]

Below is the solution using Recursion as an approach in java

<pre><code>public class NumberOfBinaryPatterns {
 static int[] bitArray = new int[]{0,1};
 public static void main(String args[])
 { 
 System.out.println("Below are the patterns\n");
 int numberOfBits=3; // It can be any value based on the requirement, i.e. 4,5, etc. We can get the input for "numberOfBits" from user on runtime to enhance this further
 drawBinaryPattern(numberOfBits,"");
 }
 private static void drawBinaryPattern(int n,String seed)
 {
 if(n==0){
 System.out.println(seed);
 return;
 } 
 for(int j=0;j&lt;bitArray.length;j++)
 {
 String temp = seed+bitArray[j];
 drawBinaryPattern(n-1,temp);
 }
 }
}
</code></pre>

blocks|key|3741300|text|要查找给定二进制字符串(模式)的所有可能排列，例如|type|unstyled|depth|inlineStyleRanges|entityRanges|data|3741301|1000的排列是1000,0100,0010,0001|3741302|void+permutation(int+no_ones,+int+no_zeroes,+string+accum){
++++if(no_ones+==+0){
++++++++for(int+i=0;i<no_zeroes;i%2B%2B){
++++++++++++accum+%2B=+"0";
++++++++}

++++++++cout+<<+accum+<<+endl;
++++++++return;
++++}
++++else+if(no_zeroes+==+0){
++++++++for(int+j=0;j<no_ones;j%2B%2B){
++++++++++++accum+%2B=+"1";
++++++++}

++++++++cout+<<+accum+<<+endl;
++++++++return;
++++}

++++permutation+(no_ones+-+1,+no_zeroes,+accum+%2B+"1");
++++permutation+(no_ones+,+no_zeroes+-+1,+accum+%2B+"0");
}

int+main(){
++++string+append+=+"";

++++//finding+permutation+of+11000+++
++++permutation(2,+6,+append);++//the+permutations+are+

++++//11000
++++//10100
++++//10010
++++//10001
++++//01100
++++//01010

++++cin.get();+
}|code-block|syntax|javascript|3741303|entityMap^0|0|0|0^^$0|@$1|2|3|4|5|6|7|K|8|@]|9|@]|A|$]]|$1|B|3|C|5|6|7|L|8|@]|9|@]|A|$]]|$1|D|3|E|5|F|7|M|8|@]|9|@]|A|$G|H]]|$1|I|3|-4|5|6|7|N|8|@]|9|@]|A|$]]]|J|$]]

To find all possible permutations of a given binary string(pattern) for example

The permutations of 1000 are 1000, 0100, 0010, 0001

<pre><code>void permutation(int no_ones, int no_zeroes, string accum){
 if(no_ones == 0){
 for(int i=0;i&lt;no_zeroes;i++){
 accum += "0";
 }

 cout &lt;&lt; accum &lt;&lt; endl;
 return;
 }
 else if(no_zeroes == 0){
 for(int j=0;j&lt;no_ones;j++){
 accum += "1";
 }

 cout &lt;&lt; accum &lt;&lt; endl;
 return;
 }

 permutation (no_ones - 1, no_zeroes, accum + "1");
 permutation (no_ones , no_zeroes - 1, accum + "0");
}

int main(){
 string append = "";

 //finding permutation of 11000 
 permutation(2, 6, append); //the permutations are 

 //11000
 //10100
 //10010
 //10001
 //01100
 //01010

 cin.get(); 
}
</code></pre>

What I am trying to do is print all the possibilities of a binary number n digits long. In other words, with a 4 digit number:

<pre><code>0001
0010
0100
1000
</code></pre>

..etc

To be honest, I have no idea of where to even start with this (other than I figure I'd need to use a loop, and probably an array) so any pointers in the right direction would be appreciated.

Print list of binary permutations

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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我想要做的是打印一个n位长的二进制数字的所有可能性。换句话说，用一个4位的数字：0001001001001000..etc老实说，我甚至不知道从哪里开始(除了我认为我需要使用一个循环，可能还有一个数组)，所以任何指向正确方向的指针都将不胜感激。

问打印二进制排列列表
EN

回答 7

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