Java 7, 541 byte
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);}
Giữ trật tự ban đầu làm tôi khó chịu trong thời gian dài, nếu không nó sẽ chỉ là một vòng lặp dễ dàng và nguyên tắc gọi đệ quy. Tuy nhiên, thử thách thú vị để tìm ra trong khi giữ trật tự.
Giải trình:
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)
Mã kiểm tra:
Hãy thử nó ở đây.
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));
}
}
Đầu ra:
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 khi độ dài là số lẻ?