问题
With n=5 and k=3 the following loop will do it
List<String> l=new ArrayList<String>();
l.add("A");l.add("B");l.add("C");l.add("D");l.add("E");
int broadcastSize = (int) Math.pow(2, l.size());
for (int i = 1; i < broadcastSize; i++) {
StringBuffer buffer = new StringBuffer(50);
int mask = i;
int j = 0;
int size=0;
System.out.println();
while (mask > 0) {
if ((mask & 1) == 1) {
System.out.println(".. "+mask);
buffer.append(l.get(j));
if (++size>3){
buffer = new StringBuffer(50);
break;
}
}
System.out.println(" "+mask);
mask >>= 1;
j++;
}
if (buffer.length()>0)
System.out.println(buffer.toString());
}
but it's not efficient I would like to do it with Banker's sequence and thus explore first singletons, then pairs, then 3-tuple and stop.
I did not find a way do that, but at least this loop should be more efficient:
List<String> l=new ArrayList<String>();
l.add("A");l.add("B");l.add("C");l.add("D");l.add("E");
int broadcastSize = (int) Math.pow(2, l.size());
for (int i = 1; i < broadcastSize; i++) {
StringBuffer buffer = new StringBuffer(50);
int mask = i;
int j = 0;
if (StringUtils.countMatches(Integer.toBinaryString(i), "1") < 4){
while (mask > 0) {
if ((mask & 1) == 1) {
buffer.append(l.get(j));
}
mask >>= 1;
j++;
}
if (buffer.length()>0)
System.out.println(buffer.toString());
}
}
there is also: but k embedded loops looks ugly
//singleton
for (int i = 0; i < l.size(); i++) {
System.out.println(l.get(i));
}
//pairs
for (int i = 0; i < l.size(); i++) {
for (int j = i+1; j < l.size(); j++) {
System.out.println(l.get(i)+l.get(j));
}
}
//3-tuple
for (int i = 0; i < l.size(); i++) {
for (int j = i+1; j < l.size(); j++) {
for (int k = j+1; k < l.size(); k++) {
System.out.println(l.get(i)+l.get(j)+l.get(k));
}
}
}
//...
// k-tuple
回答1:
This technique is called Gosper's hack. It only works for n <= 32
because it uses the bits of an int
, but you can increase it to 64 if you use a long
.
int nextCombo(int x) {
// moves to the next combination with the same number of 1 bits
int u = x & (-x);
int v = u + x;
return v + (((v ^ x) / u) >> 2);
}
...
for (int x = (1 << k) - 1; (x >>> n) == 0; x = nextCombo(x)) {
System.out.println(Integer.toBinaryString(x));
}
For n = 5
and k = 3
, this prints
111
1011
1101
1110
10011
10101
10110
11001
11010
11100
exactly as you'd expect.
回答2:
this should be the most efficient way, even if k embedded loops looks ugly
//singleton
for (int i = 0; i < l.size(); i++) {
System.out.println(l.get(i));
}
//pairs
for (int i = 0; i < l.size(); i++) {
for (int j = i+1; j < l.size(); j++) {
System.out.println(l.get(i)+l.get(j));
}
}
//3-tuple
for (int i = 0; i < l.size(); i++) {
for (int j = i+1; j < l.size(); j++) {
for (int k = j+1; k < l.size(); k++) {
System.out.println(l.get(i)+l.get(j)+l.get(k));
}
}
}
// ...
//k-tuple
回答3:
Apache commons has iterators for subsets of size k, and for permutations. Here is an iterator that iterates through 1-k tuples of an n-tuple, that combines the two:
import java.util.ArrayList;
import java.util.Arrays;
import java.util.Iterator;
import java.util.List;
import org.apache.commons.collections4.iterators.PermutationIterator;
import org.apache.commons.math3.util.Combinations;
public class AllTuplesUpToKIterator implements Iterator<List<Integer>> {
private Iterator<int[]> combinationIterator;
private PermutationIterator<Integer> permutationIterator;
int i;
int k;
int n;
public AllTuplesUpToKIterator(int n, int k) {
this.i = 1;
this.k = k;
this.n = n;
combinationIterator = new Combinations(n, 1).iterator();
permutationIterator = new PermutationIterator<Integer>(intArrayToIntegerList(combinationIterator.next()));
}
@Override
public boolean hasNext() {
if (permutationIterator.hasNext()) {
return true;
} else if (combinationIterator.hasNext()) {
return true;
} else if (i<k) {
return true;
} else {
return false;
}
}
@Override
public List<Integer> next() {
if (!permutationIterator.hasNext()) {
if (!combinationIterator.hasNext()) {
i++;
combinationIterator = new Combinations(n, i).iterator();
}
permutationIterator = new PermutationIterator<Integer>(intArrayToIntegerList(combinationIterator.next()));
}
return permutationIterator.next();
}
@Override
public void remove() {
// TODO Auto-generated method stub
}
public static List<Integer> intArrayToIntegerList(int[] arr) {
List<Integer> result = new ArrayList<Integer>();
for (int i=0; i< arr.length; i++) {
result.add(arr[i]);
}
return result;
}
public static void main(String[] args) {
int n = 4;
int k = 2;
for (AllTuplesUpToKIterator iter= new AllTuplesUpToKIterator(n, k); iter.hasNext();) {
System.out.println(iter.next());
}
}
}
来源:https://stackoverflow.com/questions/10838666/get-all-1-k-tuples-in-a-n-tuple