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Key to CSE142 Final, Autumn 2014 handout #23
1. Expression Value
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2 + 3 * (2 + 3) + 17 / 4 21
67 % 10 % 5 * 6 - 3 % 6 9
7.5 * 2 + 19 / 2 / 2.0 - 0.5 - 1 / 2 19.0
4 + 2 * 2 + "2 * 2" + 2 * 2 + 4 "82 * 244"
43 / 4 / 3 * 5.0 / 2 + 5 / 2 9.5
2. Original List Final List
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{1} {1}
{2, 5, 9} {2, 7, 9}
{2, 3, 7, 2} {2, 5, 2, 2}
{2, 4, 5, 4, 3} {2, 6, -1, 3, 3}
{5, 1, 8, 4, 8, 4} {5, 6, 2, 6, 2, 4}
3. Inheritance. The output produced is as follows.
d a d b
c 1 a 1 d 1 c 1
d 2 d 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 int analyzeParagraphs(Scanner input) {
int max = 0;
while (input.hasNextLine()) {
String line = input.nextLine();
int count = 0;
while (!line.equals("<p>")) {
count++;
line = input.nextLine();
}
System.out.println(count + "-line paragraph");
if (count > max) {
max = count;
}
}
return max;
}
6. Arrays. One possible solution appears below.
public static boolean hasAlternatingParity(int[] list) {
for (int i = 0; i < list.length - 1; i++) {
if (list[i] % 2 == list[i + 1] % 2) {
return false;
}
}
return true;
}
7. ArrayLists. One possible solution appears below.
public static void collapse(ArrayList<Integer> list) {
for (int i = 0; i < list.size() - 1; i++) {
int sum = list.get(i) + list.get(i + 1);
list.set(i, sum);
list.remove(i + 1);
}
}
8. Critters. One possible solution appears below.
public class Counter extends Critter {
private int count;
private int max;
public Counter(int n) {
count = 1;
max = n;
}
public Action getMove(CritterInfo info) {
if (count < max) {
count++;
} else {
count = 1;
}
if (info.getFront() == Neighbor.OTHER) {
return Action.INFECT;
} else if (info.getFront() == Neighbor.EMPTY) {
return Action.HOP;
} else
return Action.LEFT;
}
public Color getColor() {
return Color.BLUE;
}
public String toString() {
return count + "/" + max;
}
}
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;
}
}
]]>
Stuart Reges
Last modified: Fri Sep 16 16:16:07 PDT 2011