<![CDATA[
Key to Sample CSE143 Midterm, Summer 2009 handout #21
1. Method Call Output Produced
-----------------------------------------------
mystery(3, 3); =3=
mystery(5, 1); *
mystery(1, 5); 5 4 =3= 2 1
mystery(2, 7); 7 6 5 * 4 3 2
mystery(1, 8); 8 7 6 5 * 4 3 2 1
2. One possible solution appears below.
public String repeat(String s, int n) {
if (n < 0)
throw new IllegalArgumentException();
else if (n == 0)
return "";
else if (n == 1)
return s;
else if (n % 2 == 0) {
String temp = repeat(s, n / 2);
return temp + temp;
// alternative without temp: return repeat(s + s, n / 2);
} else
return s + repeat(s, n - 1);
}
3. One possible solution appears below.
public int maxAdjacentPairSum() {
if (front == null || front.next == null)
return 0;
else {
ListNode current = front;
int max = front.data + front.next.data;
while (current.next != null) {
int next = current.data + current.next.data;
if (next > max)
max = next;
current = current.next;
}
return max;
}
}
4. Statement Output
------------------------------------------------------------
var1.method1(); Blue 1/Blue 2
var2.method1(); Green 1
var3.method1(); compiler error
var4.method1(); Green 1
var1.method2(); Blue 2
var2.method2(); Red 2/Blue 2
var3.method2(); compiler error
var4.method2(); Red 2/Blue 2
var1.method3(); compiler error
var2.method3(); Green 3
var3.method3(); compiler error
var4.method3(); compiler error
((Blue)var3).method1(); Blue 1/White 2
((Red)var3).method2(); White 2
((White)var3).method3(); White 3
((White)var4).method3(); runtime error
((Green)var5).method3(); runtime error
((Red)var5).method1(); Blue 1/Red 2/Blue 2
((Blue)var6).method3(); compiler error
((Green)var6).method3(); runtime error
5. One possible solution appears below.
public void switchPairs(Stack<Integer> s) {
Queue<Integer> q = new LinkedQueue<Integer>();
while (!s.isEmpty())
q.enqueue(s.pop());
while (!q.isEmpty())
s.push(q.dequeue());
while (!s.isEmpty())
q.enqueue(s.pop());
while (q.size() > 1) {
int n1 = q.dequeue();
int n2 = q.dequeue();
s.push(n2);
s.push(n1);
}
if (!q.isEmpty())
s.push(q.dequeue());
}
6. One possible solution appears below.
public int maxCount() {
if (size == 0) {
return 0;
} else {
int max = 1;
int count = 1;
for (int i = 1; i < size; i++)
if (elementData[i] == elementData[i - 1]) {
count++;
if (count > max) {
max = count;
}
} else
count = 1;
return max;
}
}
]]>