Java 7, 541 Bytes
import java.util.*;List l=new ArrayList(),L=new ArrayList();String c(int n){l.add(x(n));return a(n+" ",l,n);}String a(String r,List q,Integer n){boolean e=q.equals(l),E=q.equals(L);if(e)L.clear();else l.clear();for(String i:new ArrayList<String>(q)){int s=i.length()/2,a=n.parseInt(i.substring(0,s),2),z=n.parseInt(i.substring(s),2);r+=a+" "+z+" ";if(e&a>1)L.add(x(a));if(e&z>1)L.add(x(z));if(E&a>1)l.add(x(a));if(E&z>1)l.add(x(z));}if(e&L.size()>0)r=a(r,L,n);if(E&l.size()>0)r=a(r,l,n);return r;}String x(Integer n){return n.toString(n,2);}
Die ursprüngliche Bestellung beizubehalten, hat mich über den Haufen geworfen, sonst wäre es nur eine einfache Schleife und ein rekursives Aufrufprinzip. Trotzdem ist es eine lustige Herausforderung, dies herauszufinden, während die Reihenfolge beibehalten wird.
Erläuterung:
import java.util.*; // Required import for List and Array List
List l=new ArrayList(),L=new ArrayList();
// Two Lists on class-level
String c(int n){ // Method (1) with integer parameter and String return-type
l.add(x(n)); // Start by adding the binary-String of the input integer to list `l`
return a(n+" ",l,n); // And let the magic begin in method `a` (2)
} // End of method (1)
String a(String r,List q,Integer n){ // Method (2) with a bunch of parameters and String return-type
boolean e=q.equals(l),E=q.equals(L); // Determine which of the two class-level Lists the parameter-List is
if(e) // If it's `l`:
L.clear(); // Empty `L`
else // If it's `L` instead:
l.clear(); // Empty `l`
for(String i:new ArrayList<String>(q)){
// Loop over the input list (as new ArrayList to remove the reference)
int s=i.length()/2, // Get the length of the current item in the list divided by 2
// NOTE: Java automatically floors on integer division,
// which is exactly what we want for the splitting of odd-length binary-Strings
a=n.parseInt(i.substring(0,s),2), // Split the current binary-String item in halve, and convert the first halve to an integer
z=n.parseInt(i.substring(s),2); // And do the same for the second halve
r+=a+" "+z+" "; // Append the result-String with these two integers
if(e&a>1) // If the parameter List is `l` and the first halve integer is not 0:
L.add(x(a)); // Add this integer as binary-String to list `L`
if(e&z>1) // If the parameter List is `l` and the second halve integer is not 0:
L.add(x(z)); // Add this integer as binary-String to List `L`
if(E&a>1) // If the parameter List is `L` and the first halve integer is not 0:
l.add(x(a)); // Add this integer as binary-String to List `l`
if(E&z>1) // If the parameter List is `L` and the second halve integer is not 0:
l.add(x(z)); // Add this integer as binary-String to List `l`
} // End of loop
if(e&L.size()>0) // If the parameter List is `l` and List `L` now contains any items:
r=a(r,L,n); // Recursive call with List `L` as parameter
if(E&l.size()>0) // If the parameter List is `L` and List `l` now contains any items:
r=a(r,l,n); // Recursive call with List `l` as parameter
return r; // Return the result-String with the now appended numbers
} // End of method (2)
String x(Integer n){ // Method (3) with Integer parameter and String return-type
return n.toString(n,2); // Convert the integer to its Binary-String
} // End of method (3)
Testcode:
Probieren Sie es hier aus.
import java.util.*;
class M{
List l=new ArrayList(),L=new ArrayList();String c(int n){l.add(x(n));return a(n+" ",l,n);}String a(String r,List q,Integer n){boolean e=q.equals(l),E=q.equals(L);if(e)L.clear();else l.clear();for(String i:new ArrayList<String>(q)){int s=i.length()/2,a=n.parseInt(i.substring(0,s),2),z=n.parseInt(i.substring(s),2);r+=a+" "+z+" ";if(e&a>1)L.add(x(a));if(e&z>1)L.add(x(z));if(E&a>1)l.add(x(a));if(E&z>1)l.add(x(z));}if(e&L.size()>0)r=a(r,L,n);if(E&l.size()>0)r=a(r,l,n);return r;}String x(Integer n){return n.toString(n,2);}
public static void main(String[] a){
M m=new M();
System.out.println(m.c(255));
m.l.clear();
m.L.clear();
System.out.println(m.c(225));
m.l.clear();
m.L.clear();
System.out.println(m.c(32));
}
}
Ausgabe:
255 15 15 3 3 3 3 1 1 1 1 1 1 1 1
225 14 1 3 2 1 1 1 0
32 4 0 1 0
0s aufgefüllten Binärziffern, wenn die Länge ungerade ist?