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Key to CSE142 Sample Final, Spring 2016 handout #20
1. Reference Mystery. The program produces the following output.
108 [5, 10, 14]
8 [5, 10, 14]
20 8
2. Original List Final List
--------------------------------------------------------------
{4, 1, 3} {4, 1, 2}
{2, 1, 3, 2} {2, 1, 2, 4}
{3, 6, 2, 9, 5} {3, 6, 2, 4, 6}
{1, 1, 1, 1, 8} {1, 1, 2, 1, 6}
{1, 3, 4, 6, 2, 9} {1, 3, 2, 4, 2, 8}
3. Inheritance. The output produced is as follows.
a c a c
a 1 c 1 a 1 d 1
b 2 c 2 c 2 c 2
4. Token-Based File Processing. One possible solution appears below.
public static void reportScore(Scanner input) {
int score = 0;
int count = 0;
while (input.hasNextInt()) {
int next = input.nextInt();
String type = input.next();
count += next;
if (type.equals("*")) {
score = score + next;
} else {
score = score - next;
}
}
String name = input.next();
System.out.println(name + " got " + score + " of " + count);
}
5. Line-Based File Processing. One possible solution appears below.
public static void reverseAndFlip(Scanner input) {
while (input.hasNextLine()) {
String first = input.nextLine();
if (input.hasNextLine()) {
String second = input.nextLine();
for (int i = second.length() - 1; i >= 0; i--) {
System.out.print(second.charAt(i));
}
System.out.println();
}
System.out.println(first);
}
}
6. Arrays. One possible solution appears below.
public static boolean isPairwiseSorted(int[] list) {
for (int i = 0; i < list.length - 1; i += 2) {
if (list[i] > list[i + 1]) {
return false;
}
}
return true;
}
7. ArrayLists. Two possible solutions appear below.
public static void reverse3(ArrayList<Integer> list) {
for (int i = 0; i < list.size() - 2; i += 3) {
int n1 = list.get(i);
int n3 = list.get(i + 2);
list.set(i, n3);
list.set(i + 2, n1);
}
}
public static void reverse3(ArrayList<Integer> list) {
for (int i = 0; i < list.size() - 2; i += 3) {
list.add(i, list.remove(i + 2));
list.add(i + 2, list.remove(i + 1));
}
}
8. Critters. One possible solution appears below.
public class Ferret extends Critter {
private int infectCount;
private Random r;
public Ferret() {
r = new Random();
}
public Action getMove(CritterInfo info) {
if (infectCount > 0) {
infectCount--;
}
if (info.getFront() == Neighbor.OTHER) {
infectCount = 5;
return Action.INFECT;
} else if (info.getFront() == Neighbor.EMPTY) {
return Action.HOP;
} else {
int choice = r.nextInt(2);
if (choice == 0) {
return Action.LEFT;
} else {
return Action.RIGHT;
}
}
}
public Color getColor() {
if (infectCount > 0) {
return Color.RED;
} else {
return Color.BLUE;
}
}
public String toString() {
return "I=" + infectCount;
}
}
9. Arrays. One possible solution appears below.
public static int[] splice(int[] list, int from, int to) {
int[] result = new int[list.length];
int index = 0;
for (int i = to; i < list.length; i++) {
result[index] = list[i];
index++;
}
for (int i = from; i < to; i++) {
result[index] = list[i];
index++;
}
for (int i = 0; i < from; i++) {
result[index] = list[i];
index++;
}
return result;
}
10. Programming. One possible solution appears below.
public static boolean isMatch(String pattern, String text) {
int j = 0;
for (int i = 0; i < pattern.length(); i++) {
char ch1 = pattern.charAt(i);
if (ch1 == '*') {
int diff = text.length() - pattern.length() + 1;
if (diff < 0) {
return false;
} else {
j += diff;
}
} else {
if (j >= text.length()) {
return false;
}
char ch2 = text.charAt(j);
j++;
if (ch1 != '.' && ch1 != ch2) {
return false;
}
}
}
return j == text.length();
}
below is a solution to the 6-point problem:
public static boolean isMatch(String pattern, String text) {
if (text.length() != pattern.length()) {
return false;
}
for (int i = 0; i < pattern.length(); i++) {
char ch1 = pattern.charAt(i);
char ch2 = text.charAt(i);
if (ch1 != '.' && ch1 != ch2) {
return false;
}
}
return true;
}
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