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Key to CSE142 Final, Winter 2019 handout #22
1. Reference Mystery. The program produces the following output.
3 [3, 6, 8]
7 [0, 3, 8]
7 7
2. Original List Final List
--------------------------------------------------------------
{0, 2, 3} {0, 2, 5}
{1, 1, 4, 5} {1, 1, 3, 7}
{3, 1, 2, 9, 5} {3, 1, 3, 7, 5}
{5, 1, 1, 1, 8} {5, 1, 3, 7, 8}
{2, 1, 1, 2, 3, 4} {2, 1, 3, 7, 3, 7}
3. Inheritance. The output produced is as follows.
d a d d
b 1 b 1 d 1 c 1
b 2 b 2 d 2 d 2
4. Token-Based File Processing. One possible solution appears below.
public static double evaluate(Scanner input) {
double result = input.nextDouble();
while (input.hasNext()) {
String operator = input.next();
double num = input.nextDouble();
if (operator.equals("+")) {
result += num;
} else { // operator.equals("-")
result -= num;
}
}
return result;
}
5. Line-Based File Processing. One possible solution appears below.
public static void reportBlankLines(Scanner input) {
int line = 0;
int count = 0;
while (input.hasNextLine()) {
String text = input.nextLine();
line++;
if (text.length() == 0) {
System.out.println("line " + line + " is blank");
count++;
}
}
System.out.println("total blank lines = " + count);
}
6. Arrays. One possible solution appears below.
public static void switchPairs(int[] list) {
for (int i = 0; i < list.length - 1; i += 2) {
int temp = list[i];
list[i] = list[i + 1];
list[i + 1] = temp;
}
}
7. ArrayLists. One possible solution appears below.
public static void removeAdjacentMatches(ArrayList<Integer> list) {
for (int i = 0; i < list.size() - 1; i++) {
if (list.get(i) == list.get(i + 1)) {
list.remove(i + 1);
i--;
}
}
}
8. Critters. One possible solution appears below.
public class Iguana extends Critter {
private Random r;
private String display;
private int count;
public Iguana() {
r = new Random();
display = "<->";
}
public Action getMove(CritterInfo info) {
count++;
if (info.getFront() == Neighbor.OTHER) {
display = "<I>";
return Action.INFECT;
} else {
int flip = r.nextInt(2);
if (flip == 0) {
display = "<L>";
return Action.LEFT;
} else {
display = "<R>";
return Action.RIGHT;
}
}
}
public Color getColor() {
if (count % 2 == 0) {
return Color.BLUE;
} else {
return Color.RED;
}
}
public String toString() {
return display;
}
}
9. Arrays. One possible solution appears below.
public static int[] combineSublists(int[] list1, int[] list2,
int start, int stop) {
int sublistLength = stop - start;
int[] result = new int[2 * sublistLength];
for (int i = start; i < stop; i++) {
result[i - start] = list1[i];
result[i - start + sublistLength] = list2[i];
}
return result;
}
10. Programming. Two possible solutions appear below.
public static void removeZeros(int[] list) {
for (int i = list.length - 2; i >= 0; i--) {
if (list[i] == 0) {
for (int j = i; j < list.length - 1; j++) {
list[j] = list[j + 1];
}
list[list.length - 1] = 0;
}
}
}
public static void removeZeros(int[] list) {
int numNonZero = 0;
for (int i = 0; i < list.length; i++) {
if (list[i] != 0) {
list[numNonZero] = list[i];
numNonZero++;
}
}
for (int i = numNonZero; i < list.length; i++) {
list[i] = 0;
}
}
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