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#415 disjoint sets impl #421
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Original file line number | Diff line number | Diff line change |
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@@ -26,7 +26,6 @@ | |
import java.util.Set; | ||
import org.cactoos.list.ListOf; | ||
import org.hamcrest.core.AllOf; | ||
import org.junit.jupiter.api.Disabled; | ||
import org.junit.jupiter.api.Test; | ||
import org.llorllale.cactoos.matchers.Assertion; | ||
import org.llorllale.cactoos.matchers.HasValuesMatching; | ||
|
@@ -38,7 +37,6 @@ | |
public final class DisjointTest { | ||
There was a problem hiding this comment. Choose a reason for hiding this commentThe reason will be displayed to describe this comment to others. Learn more. @HDouss Could you please add one more test where the nodes in the graph are not connected at all? |
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|
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@Test | ||
@Disabled | ||
@SuppressWarnings("unchecked") | ||
public void calculatesDisjointSets() throws Exception { | ||
final Node one = new Node.Simple("one"); | ||
|
@@ -76,4 +74,36 @@ public void calculatesDisjointSets() throws Exception { | |
).affirm(); | ||
} | ||
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@Test | ||
@SuppressWarnings("unchecked") | ||
public void calculatesDisjointSetsForUnconnected() throws Exception { | ||
final Node one = new Node.Simple("1"); | ||
final Node two = new Node.Simple("2"); | ||
final Node three = new Node.Simple("3"); | ||
final Node four = new Node.Simple("4"); | ||
final Graph graph = new FakeGraph( | ||
new ListOf<>(one, two, three, four) | ||
); | ||
new Assertion<>( | ||
"Must build disjoint sets for unconnected graph", | ||
new Disjoint(graph).value(), | ||
new AllOf<Iterable<Set<Node>>>( | ||
new ListOf<>( | ||
new HasValuesMatching<>( | ||
set -> set.contains(one) && set.size() == 1 | ||
), | ||
new HasValuesMatching<>( | ||
set -> set.contains(two) && set.size() == 1 | ||
), | ||
new HasValuesMatching<>( | ||
set -> set.contains(three) && set.size() == 1 | ||
), | ||
new HasValuesMatching<>( | ||
set -> set.contains(four) && set.size() == 1 | ||
) | ||
) | ||
) | ||
).affirm(); | ||
} | ||
|
||
} |
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@HDouss Sorry, I could catch the whole algorithm, but is it possible to use
iterator.hasNext()
for theunvisited
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@fanifieiev it is not really possible, unless to get a new iterator from the set in each iteration. This set is getting shrinked (by more than one element) in each iteration. Even the element at which the iterator is pointing is being removed from the set.