15: COLLECTIONS AND MAPS in Java

Generating QR Code in Java 15: COLLECTIONS AND MAPS
CHAPTER 15: COLLECTIONS AND MAPS
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Program output:
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Serving meals from a HashSet (order can vary): dinner breakfast lunch Serving meals from a LinkedHashSet (always insertion order): breakfast lunch dinner
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Example 1516 demonstrates set operations It determines the following relationships between two sets of characters: whether they are disjunct, that is, have no elements in common whether they have the same elements, that is, are equivalent whether one is a subset of the other whether one is a superset of the other whether they have a common subset Given a list of words as program arguments, each argument is turned into a set of characters This character set is compared with the set of all characters encountered so far in previous arguments The set encountered created at (1) accumulates characters as each argument is processed For each argument, an empty set of characters is created at (2) This characters set is populated with the characters of the current argument at (3) The program first determines if there is a common subset between the two sets at (4), ie, whether the current argument has any characters that were in previous arguments:
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// Determine whether a common subset exists (4) Set<Character> commonSubset = new HashSet<Character>(encountered); commonSubsetretainAll(characters); boolean areDisjunct = commonSubsetsize()==0;
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Note that the retainAll() operation is destructive The code at (4) does not affect the encountered and the characters sets If the size of the common subset is zero, the sets are disjunct; otherwise, the relationship must be narrowed down The subset and superset relations are determined at (5) using the containsAll() method
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// Determine superset and subset relations boolean isSubset = encounteredcontainsAll(characters); boolean isSuperset = characterscontainsAll(encountered); (5)
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The sets are equivalent if both the previous relations are true If the relations are both false, that is, no subset or superset relationship exists, the sets only have the
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154: SETS
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subset computed at (4) in common The encountered set is updated with the contents of the characters set to accumulate all characters encountered so far The addAll() method is used for this purpose at (6):
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encounteredaddAll(characters); // (6)
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As we can see from the output, the program prints the contents of the sets in the standard textual representation for collections
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Example 1516 Using Sets import javautilHashSet; import javautilSet; public class CharacterSets { public static void main(String[] args) { int numArgs = argslength; // A set keeping track of all characters previously encountered Set<Character> encountered = new HashSet<Character>(); // (1) // For each program argument in the command line for (String argument : args) { // Convert the current argument to a set of characters Set<Character> characters = new HashSet<Character>(); int size = argumentlength(); // For each character in the argument for (int j=0; j<size; j++) // add character to the characters set charactersadd(argumentcharAt(j));
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// (2)
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// (3)
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// Determine whether a common subset exists (4) Set<Character> commonSubset = new HashSet<Character>(encountered); commonSubsetretainAll(characters); boolean areDisjunct = commonSubsetsize()==0; if (areDisjunct) { Systemoutprintln(characters + " and " + encountered + " are disjunct"); } else { // Determine superset and subset relations (5) boolean isSubset = encounteredcontainsAll(characters); boolean isSuperset = characterscontainsAll(encountered); if (isSubset && isSuperset) Systemoutprintln(characters + " is equivalent to " + encountered); else if (isSubset) Systemoutprintln(characters + " is a subset of " + encountered); else if (isSuperset) Systemoutprintln(characters + " is a superset of " + encountered); else Systemoutprintln(characters + " and " + encountered + " have " + commonSubset + " in common"); }
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CHAPTER 15: COLLECTIONS AND MAPS // Update the set of characters encountered so far encounteredaddAll(characters); } } }
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// (6)
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Running the program with the following arguments:
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>java CharacterSets i said i am maids
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results in the following output:
[i] [d, [i] [a, [d, and [] are disjunct a, s, i] is a superset of [i] is a subset of [d, a, s, i] m] and [d, a, s, i] have [a] in common a, s, m, i] is equivalent to [d, a, s, m, i]