SETS AND MAPS - Introduction to Java Programming: Comprehensive Version - page 801

30
31 System.out.println( "\nIs Taipei in set2? "
32 + set2.contains( "Taipei" ));
33
34 set1.addAll(set2);
35 System.out.println( "\nAfter adding set2 to set1, set1 is "
36 + set1);
37
38 set1.removeAll(set2);
39 System.out.println( "After removing set2 from set1, set1 is "
40 + set1);
41
42 set1.retainAll(set2);
43 System.out.println( "After removing common elements in set2 "
44 + "from set1, set1 is " + set1);
45 }
46 }

contains element?

addAll

removeAll

retainAll

set1 is [San Francisco, New York, Paris, Beijing, London]
5 elements in set1

set1 is [San Francisco, New York, Paris, Beijing]
4 elements in set1

set2 is [Shanghai, Paris, London]
3 elements in set2

Is Taipei in set2? false

After adding set2 to set1, set1 is
[San Francisco, New York, Shanghai, Paris, Beijing, London]

After removing set2 from set1, set1 is
[San Francisco, New York, Beijing]

After removing common elements in set2 from set1, set1 is []

The program creates two sets (lines 4, 22). The size() method returns the number of the
elements in a set (line 14). Line 17

set1.remove( "London" );

removes London from set1 .
The contains method (line 32) checks whether an element is in the set.
Line 34

set1.addAll(set2);

adds set2 to set1 . Therefore, set1 becomes [San Francisco, New York, Shanghai,
Paris, Beijing, London] .
Line 38

set1.removeAll(set2);

removes set2 from set1 . Thus, set1 becomes [San Francisco, New York, Beijing] .
Line 42

set1.retainAll(set2);