二进制表示数组生成解决思路

二进制表示数组生成
想得到这样的数字比如输入3，得到类似下面的数组
000
001
010
011
100
101
110
111

[解决办法]
2的三次方
然后用
Integer.toBinaryString(int i)
就行了
[解决办法]

Java code

    public static void main(String[] args) {        int a = 0;        String align = "";        for (int i = 0; i < 6; i++) {            a += 1 << i;            align += "0";        }                String[] r = new String[a + 1];        for (int i = 0; i <= a; ++i) {            String t = Integer.toBinaryString(i);            String s = (align + t);            s = s.substring(t.length());            System.out.println(s);        }    }
[解决办法]
Java code
import java.text.DecimalFormat; 
 import java.util.Scanner; 
  
 public class Test_2 { 
 static public void print(double i){ 
 String[] init=new String[(int)Math.pow(2.0, i)]; 
  DecimalFormat df=new DecimalFormat("000"); 
 for(int a=0;a <init.length;a++){ 
 init[a]=df.format(Double.parseDouble(Integer.toBinaryString(a))); 
 System.out.println(init[a]); 
 } 
 } 
 public static void main(String[] args) { 
 Scanner sca=new Scanner(System.in); 
 double i=sca.nextDouble(); 
 print(i); 
  
 } 
  
 } 
 

输入3的时候的测试结果：
000
001
010
011
100
101
110
111

[解决办法]
Java code
public class Test {    public static void main(String[] args) {        String[] strs = getBinaries(3);        for(int i = 0; i < strs.length; i++) {            System.out.println(strs[i]);        }    }        public static String[] getBinaries(int len) {        if(len < 0) {            throw new IllegalArgumentException("len must be positive number");        }        if(len > 30) {            throw new IllegalArgumentException("len is too many");        }        char[] chs = new char[len];        int count = 1 << len;        String[] strs = new String[count];        for(int i = 0; i < count; i++) {            strs[i] = getBinary(i, chs);        }        return strs;    }        private static String getBinary(long num, char[] chs) {        for(int i = 0, k = chs.length - 1; i <= k; i++) {                        chs[k - i] = (char)((num & 1) + '0');            num >>= 1;        }        return new String(chs);    }}
[解决办法]
Java code
public class Test {    public static void main(String[] args) {        int value = 3;        int maxValue = (int)Math.pow(2, 3);        String[] result = new String[maxValue];        while(maxValue > 0) {            result[maxValue -- - 1] = String.format("%3s", Integer.toBinaryString(maxValue)).replaceAll(" ", "0");        }        for(int i = 0; i < result.length; i ++) {            System.out.println(result[i]);        }    }}
[解决办法]
来个简单一点的：
Java code
import java.text.DecimalFormat;public class Gen{    public static String genStr(int len)    {        for (int i = 0; i < (1 << len); i++)        {            System.out.println(new DecimalFormat("000").format(Double.parseDouble(Integer.toBinaryString(i))));        }        return null;    }    public static void main(String[] args)    {        Gen.genStr(3);    }} 
 
[解决办法]
Java code
import java.text.DecimalFormat;public class Gen{    public static String genStr(int len)    {        StringBuilder sb = new StringBuilder();        for (int i = 0; i < len; i++)        {            sb.append("0");        }        for (int i = 0; i < (1 << len); i++)        {            System.out.println(new DecimalFormat(sb.toString()).format(Double.parseDouble(Integer.toBinaryString(i))));        }        return null;    }    public static void main(String[] args)    {        Gen.genStr(3);    }}
[解决办法]
Java code
import java.text.*;public class FindPrime {    public static void toString(int i){        int max = (int)Math.pow(2, i);                String str = Integer.toBinaryString(max);                int maxLength = str.length();                StringBuffer sb = new StringBuffer();                for(int j=0; j<maxLength-1; j++)            sb.append("0");                DecimalFormat df = new DecimalFormat(sb.toString());                for(int j=0; j<max; j++)            System.out.println(df.format(Integer.parseInt(Integer.toBinaryString(j))));            }    public static void main(String args[]){        toString(3);            }}