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Answers to Questions and Exercises:
Questions
-
Question:
This lesson mentions three ways to traverse a
List
. Describe them, and note the limitations of
each.
Answer:
- Use the enhanced
for
statement:
List<Thing> list;
...
for (Thing thing : list) {
...
}
Limitations: cannot be used to add, remove, or modify elements.
- Use the traditional
for
statement together with an
Iterator
:
List<Thing> list;
...
for (Iterator<Thing> it = list.iterator(); it.hasNext(); ) {
Thing thing = it.next();
...
}
Limitations: cannot be used to modify elements.
- Use the traditional
for
statement together with an
ListIterator
:
List<Thing> list;
...
for (ListIterator<Thing> it = list.iterator(); it.hasNext(); ) {
Thing thing = it.next();
...
}
Limitations: none.
-
Question:
Consider the four core interfaces,
Set
,
List
, Queue
, and Map
.
For each of the following four assignments, specify which of the
four core interfaces is best-suited, and explain how to use it
to implement the assignment.
Answer:
- Whimsical Toys Inc (WTI) needs to record the names of
all its employees. Every month, an employee will be chosen
at random from these records to receive a free
toy.
Use a List
. Choose a random employee by
picking a number between 0
and
size()-1
.
- WTI has decided that each new product will be
named after an employee — but only first names
will be used, and each name will be used only once.
Prepare a list of unique first names.
Use a Set
. Collections that
implement this interface don't allow the same element
to be entered more than once.
- WTI decides that it only wants to use the most
popular names for its toys. Count up the number of
employees who have each first name.
Use a Map
, where the keys are first
names, and each value is a count of the number of
employees with that first name.
- WTI acquires season tickets for the local lacrosse
team, to be shared by employees. Create a waiting list
for this popular sport.
Use a Queue
. Invoke
add()
to add employees to the waiting
list, and remove()
to remove them.
-
Question:
The following program is supposed to print the string "Blue". Instead,
it throws an error. Why?
import java.util.*;
public class SortMe {
public static void main(String args[]) {
SortedSet<StringBuffer> s = new TreeSet<StringBuffer>();
s.add(new StringBuffer("Red"));
s.add(new StringBuffer("White"));
s.add(new StringBuffer("Blue"));
System.out.println(s.first());
}
}
Answer:
TreeSort
elements must be instances of a class that
implements
Comparable
. StringBuffer
does not.
Exercises
- Exercise:
Write a program that prints its arguments in random order. Do not
make a copy of the argument array.
Answer:
import java.util.*;
public class Ran {
public static void main(String[] args) {
List<String> argList = Arrays.asList(args);
Collections.shuffle(argList);
for (String arg: argList) {
System.out.format("%s ", arg);
}
System.out.println();
}
}
- Exercise:
Take the
FindDups example
and modify it to use a SortedSet
instead of a
Set
. Specify a Comparator
so that case is
ignored when sorting and identifying set elements.
Answer:
import java.util.*;
public class FindDups {
public static void main(String[] args) {
Comparator<String> comparator = new Comparator<String>() {
public int compare (String s1, String s2) {
return s1.compareToIgnoreCase(s2);
}
};
SortedSet<String> s = new TreeSet<String>(comparator);
for (String a : args)
if (!s.add(a))
System.out.println("Duplicate detected: " + a);
System.out.println(s.size() + " distinct words: " + s);
}
}
- Exercise:
Write a method that takes a
List<String>
and
applies
String.trim
to each element. To do this, you'll need to pick one of the three
iteration idioms that you described in Question 1. Two of these will
not give the result you want, so be sure to write a program that
demonstrates that the method actually works!
Answer:
The enhanced for
statement does not allow you to modify
the List
. Using an Iterator
allows you to
delete elements, but not replace an existing element or add a new one.
That leaves ListIterator
:
import java.util.*;
public class ListTrim {
static void listTrim(List<String> strings) {
for (ListIterator<String> lit = strings.listIterator();
lit.hasNext(); ) {
lit.set(lit.next().trim());
}
}
public static void main(String[] args) {
List<String> l = Arrays.asList(" red ", " white ", " blue ");
listTrim(l);
for (String s : l) {
System.out.format("\"%s\"%n", s);
}
}
}