Ten kod wyprowadza zmiany, a następnie uruchamia permutacje dla każdego unikalnego zestawu 3.
tj. Dla "A", "B", "C", "D" możliwościami są [[A, B, C], [A, B, D], [A, C, D], [B, PŁYTA CD]]. Następnie obliczamy permutacje na każdym trójkącie (lub n-tym) i dołączamy możliwości do listy.
PermutationsOfN.processSubsets (listę zestawu int k) zwrócone: [[A, B, C], [A, B, D], [A, C, D], [B, C, D ]]
Trochę dalej PermutationsOfN.permutacje (lista list, rozmiar int) zwraca:
[[A, B, C], [A, C, B], [C, A, B], [C, B, A], [B, C , A], [B, A, C], [A, B, D], [A, D, B], [D, A, B], [D, B, A], [B, D, A ], [B, A, D], [A, C, D], [A, D, C], [D, A, C], [D, C, A], [C, D, A], [C, A, D], [B, C, D], [B, D, C], [D, B, C], [D, C, B], [C, D, B], [C , B, D]]
import java.util.Collection;
import java.util.List;
import com.google.common.collect.Collections2;
import com.google.common.collect.ImmutableList;
import com.google.common.collect.Lists;
public class PermutationsOfN<T> {
public static void main(String[] args) {
List<String> f = Lists.newArrayList("A", "B", "C", "D");
PermutationsOfN<String> g = new PermutationsOfN<String>();
System.out.println(String.format("n=1 subsets %s", g.processSubsets(f, 1)));
System.out.println(String.format("n=1 permutations %s", g.permutations(f, 1)));
System.out.println(String.format("n=2 subsets %s", g.processSubsets(f, 2)));
System.out.println(String.format("n=2 permutations %s", g.permutations(f, 2)));
System.out.println(String.format("n=3 subsets %s", g.processSubsets(f, 3)));
System.out.println(String.format("n=3 permutations %s", g.permutations(f, 3)));
System.out.println(String.format("n=4 subsets %s", g.processSubsets(f, 4)));
System.out.println(String.format("n=4 permutations %s", g.permutations(f, 4)));
System.out.println(String.format("n=5 subsets %s", g.processSubsets(f, 5)));
System.out.println(String.format("n=5 permutations %s", g.permutations(f, 5)));
}
public List<List<T>> processSubsets(List<T> set, int k) {
if (k > set.size()) {
k = set.size();
}
List<List<T>> result = Lists.newArrayList();
List<T> subset = Lists.newArrayListWithCapacity(k);
for (int i = 0; i < k; i++) {
subset.add(null);
}
return processLargerSubsets(result, set, subset, 0, 0);
}
private List<List<T>> processLargerSubsets(List<List<T>> result, List<T> set, List<T> subset, int subsetSize, int nextIndex) {
if (subsetSize == subset.size()) {
result.add(ImmutableList.copyOf(subset));
} else {
for (int j = nextIndex; j < set.size(); j++) {
subset.set(subsetSize, set.get(j));
processLargerSubsets(result, set, subset, subsetSize + 1, j + 1);
}
}
return result;
}
public Collection<List<T>> permutations(List<T> list, int size) {
Collection<List<T>> all = Lists.newArrayList();
if (list.size() < size) {
size = list.size();
}
if (list.size() == size) {
all.addAll(Collections2.permutations(list));
} else {
for (List<T> p : processSubsets(list, size)) {
all.addAll(Collections2.permutations(p));
}
}
return all;
}
}
wyróżnienie trafia do meriton którego odpowiedź here pomógł mi się dogadać.
'Kolekcje2.permutations (coll) .filter (l -> l.size() == k)' – jpaugh