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Key to CSE143 Midterm, Winter 2005 handout #21
1. Method Call Value Returned
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mystery(1, 3) 9
mystery(4, 4) 7
mystery(3, 5) 21
mystery(1, 5) 25
mystery(4, 7) 40
2. One possible solution appears below.
public static void countToBy(int n, int m) {
if (m <= 0 || n <= 0)
throw new IllegalArgumentException();
else if (n <= m)
System.out.print(n);
else {
countToBy(n - m, m);
System.out.print(", " + n);
}
}
3. One possible solution appears below.
public int lastIndexOf(int n) {
int result = -1;
int index = 0;
ListNode current = this.front;
while (current != null) {
if (current.data == n) {
result = index;
}
index++;
current = current.next;
}
return result;
}
4. Statement Output
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var1.method2(); Ocean 2
var2.method2(); Pond 2
var3.method2(); Pond 2
var4.method2(); compiler error
var5.method2(); Bay 2
var6.method2(); Ocean 2
var1.method3(); Lake 3/Ocean 2
var2.method3(); compiler error
var3.method3(); compiler error
var4.method3(); compiler error
var5.method3(); Lake 3/Bay 2
var6.method3(); Lake 3/Ocean 2
((Ocean)var5).method1(); runtime error
((Lake)var3).method3(); Lake 3/Pond 2
((Lake)var4).method1(); compiler error
((Ocean)var1).method1(); Bay 1/Pond 2
((Bay)var4).method1(); Bay 1/Pond 2
((Lake)var2).method3(); runtime error
((Ocean)var5).method1(); runtime error
((Pond)var4).method2(); Bay 2
5. Two possible solutions appear below.
public boolean isPalindrome(Queue<Integer> q) {
Stack<Integer> s = new ArrayStack<Integer>();
for (int i = 0; i < q.size(); i++) {
int n = q.dequeue();
q.enqueue(n);
s.push(n);
}
boolean ok = true;
for (int i = 0; i < q.size(); i++) {
int n1 = q.dequeue();
int n2 = s.pop();
if (n1 != n2)
ok = false;
q.enqueue(n1);
}
return ok;
}
public static boolean isPalindrome(Queue<Integer> q) {
Stack<Integer> s = new ArrayStack<Integer>();
int oldSize = q.size();
for (int i = 0; i < oldSize / 2; i++) {
int n = q.dequeue();
s.push(n);
}
int oddValue = 0;
if (oldSize % 2 == 1)
oddValue = q.dequeue();
boolean ok = true;
while (!s.isEmpty()) {
int n1 = q.dequeue();
int n2 = s.pop();
if (n1 != n2)
ok = false;
q.enqueue(n1);
q.enqueue(n2);
}
for (int i = 0; i < oldSize / 2; i++) {
int n1 = q.dequeue();
int n2 = q.dequeue();
q.enqueue(n1);
s.push(n2);
}
while (!s.isEmpty())
q.enqueue(s.pop());
if (oldSize % 2 == 1)
q.enqueue(oddValue);
for (int i = 0; i < oldSize / 2; i++)
q.enqueue(q.dequeue());
return ok;
}
6. One possible solution appears below.
public boolean isPairwiseSorted() {
for (int i = 0; i < size - 1; i += 2)
if (elementData[i] > elementData[i + 1])
return false;
return true;
}
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