<![CDATA[
CSE143 Sample Midterm handout #16
Fall 2012
1. Recursive Tracing, 15 points. Consider the following method:
public void mystery(int n) {
if (n == 1) {
System.out.print(n);
} else {
System.out.print(n + ", ");
if (n % 2 == 0) {
mystery(n / 2);
} else {
mystery(3 * n + 1);
}
}
}
For each call below, indicate what output is produced:
Method Call Output Produced
mystery(1); ___________________________________________
mystery(2); ____________________________________________
mystery(4); ____________________________________________
mystery(8); ____________________________________________
mystery(10); ____________________________________________
2. Recursive Programming, 15 points. Write a recursive method called
printDashed that takes an integer as a parameter and that prints the integer
with dashes in between the digits.
The table below shows sample calls and the output that should be produced:
Method call Output Method call Output
------------------------------ ---------------------------------
printDashed(-834) -8-3-4 printDashed(6) 6
printDashed(-17) -1-7 printDashed(42) 4-2
printDashed(-4) -4 printDashed(983) 9-8-3
printDashed(0) 0 printDashed(29348) 2-9-3-4-8
Notice that no dashes are printed for positive one-digit numbers and that a
leading dash is printed only for negative numbers. You are not allowed to
construct any structured objects (no array, ArrayList, String,
StringBuilder, etc) and you may not use a while loop, for loop, or do/while
loop to solve this problem; you must use recursion.
3. Linked Lists, 15 points. Fill in the "code" column in the following table
providing a solution that will turn the "before" picture into the "after"
picture by modifying links between the nodes shown. You are not allowed to
change any existing node's data field value.
You are writing code for the ListNode class discussed in lecture:
public class ListNode {
public int data; // data stored in this node
public ListNode next; // link to next node in the list
<constructors>
}
As in the lecture examples, all lists are terminated by null.
before after code
+-----------------------+-----------------------+-----------------------------
| | | |
| | | |
| p->[1]->[2] | p->[2] | |
| | | |
| | | |
| q->[3]->[4] | q->[3]->[4]->[1] | |
| | | |
| | | |
+-----------------------+-----------------------+-----------------------------+
| | | |
| | | |
| p->[1]->[2] | p->[2]->[1] | |
| | | |
| | | |
| q->[3]->[4] | q->[4]->[3] | |
| | | |
| | | |
+-----------------------+-----------------------+-----------------------------+
| | | |
| | | |
| p->[1]->[2]->[3] | p->[1]->[4] | |
| | | |
| | | |
| q->[4] | q->[3]->[2] | |
| | | |
| | | |
+-----------------------+-----------------------+-----------------------------+
4. Details of inheritance, 20 points. Assuming that the following classes have
been defined:
public class Gorge extends Cliff {
public void method2() {
System.out.println("Gorge 2");
}
public void method3() {
System.out.println("Gorge 3");
}
}
public class Hill extends Peak {
public void method2() {
System.out.println("Hill 2");
}
public void method3() {
System.out.println("Hill 3");
}
}
public class Peak {
public void method1() {
System.out.println("Peak 1");
method3();
}
public void method3() {
System.out.println("Peak 3");
}
}
public class Cliff extends Peak {
public void method3() {
System.out.println("Cliff 3");
super.method3();
}
}
And assuming the following variables have been defined:
Peak var1 = new Cliff();
Gorge var2 = new Gorge();
Peak var3 = new Hill();
Peak var4 = new Gorge();
Peak var5 = new Peak();
Object var6 = new Cliff();
In the table below, indicate in the right-hand column the output produced by
the statement in the left-hand column. If the statement produces more than
one line of output, indicate the line breaks with slashes as in "a/b/c" to
indicate three lines of output with "a" followed by "b" followed by "c". If
the statement causes an error, fill in the right-hand column with either the
phrase "compiler error" or "runtime error" to indicate when the error would
be detected.
Statement Output
------------------------------------------------------------
var1.method1(); ____________________________
var2.method1(); ____________________________
var3.method1(); ____________________________
var4.method1(); ____________________________
var5.method1(); ____________________________
var6.method1(); ____________________________
var1.method2(); ____________________________
var2.method2(); ____________________________
var3.method2(); ____________________________
var1.method3(); ____________________________
var2.method3(); ____________________________
var3.method3(); ____________________________
((Gorge)var6).method1(); ____________________________
((Cliff)var3).method2(); ____________________________
((Gorge)var4).method2(); ____________________________
((Gorge)var3).method2(); ____________________________
((Hill)var3).method2(); ____________________________
((Gorge)var1).method1(); ____________________________
((Cliff)var4).method3(); ____________________________
((Peak)var6).method3(); ____________________________
5. Stacks/Queues, 25 points. Write a method called removeMin that takes a
stack of integers as a parameter and that removes and returns the smallest
value from the stack. For example, if a variable called s stores the
following sequence of values:
bottom [2, 8, 3, 19, 7, 3, 2, 42, 9, 3, 2, 7, 12, -8, 4] top
and you make the following call:
int n = removeMin(s);
the method removes and returns the value -8 from the stack, so that the
variable n will be -8 after the call and s will store the following values:
bottom [2, 8, 3, 19, 7, 3, 2, 42, 9, 3, 2, 7, 12, 4] top
If the minimum value appears more than once, all occurrences of the minimum
should be removed from the stack. For example, given the ending value of
the stack above, if we again call removeMin(s), the method would return 2
and would leave the stack in the following state:
bottom [8, 3, 19, 7, 3, 42, 9, 3, 7, 12, 4] top
You are to use one queue as auxiliary storage to solve this problem. You
may not use any other auxiliary data structures to solve this problem,
although you can have as many simple variables as you like. You also may
not solve the problem recursively. Your solution must run in O(n) time
where n is the size of the stack. Use the Stack and Queue structures
described in the cheat sheet and obey the restrictions described there.
6. Array Programming, 10 points. Write a method called mirror that doubles the
size of a list of integers by appending the mirror image of the original
sequence to the end of the list. The mirror image is the same sequence of
values in reverse order. For example, if a variable called list stores this
sequence of values:
[1, 3, 2, 7]
and you make the following call:
list.mirror();
then it should store the following values after the call:
[1, 3, 2, 7, 7, 2, 3, 1]
Notice that it has been doubled in size by having the original sequence
appearing in reverse order at the end of the list.
You are writing a method for the ArrayIntList class discussed in lecture:
public class ArrayIntList {
private int[] elementData; // list of integers
private int size; // current # of elements in the list
<methods>
}
You are not to call any other ArrayIntList methods to solve this problem,
you are not allowed to define any auxiliary data structures (no array,
ArrayList, etc). You may assume that the array has sufficient capacity to
store the new values.
]]>