生成二进制序列
Generating Binary Sequences
我想生成所有可能的二进制数字序列,其中列表中的每个序列都限制为特定数量的 1,并且填充零以使每个列表的长度相同。
例如,如果序列应该是 4 个数字长,并且有 2 个一,则所有序列将是:
1100 1010 1001 0110 0101 0011
并且保留数字前面的零。
试试这个:
//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++){
//Make 16 bit long binary Strings
String s = String.format("%16s", 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());
}
如果要保留 0,则只需添加填充:
int end = 100; //Change this
for (int i = 0; i <= end; i++) {
String bytestring = Integer.toBinaryString(i);
String padding = "00000000000000000000000000000000";
bytestring = padding.substring(0, 32 - bytestring.length()) + bytestring;
System.out.println(bytestring);
}
这可以使用递归函数调用来解决:
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 + "1");
} else {
results.add(prefix + "0");
}
} else {
checkSubString(numCount-1, oneCount , prefix + "0");
checkSubString(numCount-1, oneCount-1, prefix + "1");
}
}
}
}
需要 org.apache.commons.lang.StringUtils
,但这使它变得很短:
final int digits = 4;
final int onesrequired = 2;
int maxindex = (int) Math.pow(2, digits);
for (int i = 0; i < maxindex; i++) {
String binaryStr = Integer.toBinaryString(i);
if (StringUtils.countMatches(binaryStr, "1") == onesrequired) {
System.out.print(String.format("%" + digits + "s", binaryStr).replace(' ', '0') + " ");
}
}
我想生成所有可能的二进制数字序列,其中列表中的每个序列都限制为特定数量的 1,并且填充零以使每个列表的长度相同。
例如,如果序列应该是 4 个数字长,并且有 2 个一,则所有序列将是:
1100 1010 1001 0110 0101 0011
并且保留数字前面的零。
试试这个:
//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++){
//Make 16 bit long binary Strings
String s = String.format("%16s", 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());
}
如果要保留 0,则只需添加填充:
int end = 100; //Change this
for (int i = 0; i <= end; i++) {
String bytestring = Integer.toBinaryString(i);
String padding = "00000000000000000000000000000000";
bytestring = padding.substring(0, 32 - bytestring.length()) + bytestring;
System.out.println(bytestring);
}
这可以使用递归函数调用来解决:
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 + "1");
} else {
results.add(prefix + "0");
}
} else {
checkSubString(numCount-1, oneCount , prefix + "0");
checkSubString(numCount-1, oneCount-1, prefix + "1");
}
}
}
}
需要 org.apache.commons.lang.StringUtils
,但这使它变得很短:
final int digits = 4;
final int onesrequired = 2;
int maxindex = (int) Math.pow(2, digits);
for (int i = 0; i < maxindex; i++) {
String binaryStr = Integer.toBinaryString(i);
if (StringUtils.countMatches(binaryStr, "1") == onesrequired) {
System.out.print(String.format("%" + digits + "s", binaryStr).replace(' ', '0') + " ");
}
}