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Key to CSE142 Final handout #24
Winter 2011
1. Expression Value
---------------------------------------------------
(6 + 7) / 4 / 2.0 1.5
13 / 2 * 3 % 10 - 1 7
30 / (2 * 6) + 1.5 3.5
13 % 5 + 5 * 3 / 4 6
"3 * 4" + 3 * 4 + 10 "3 * 41210"
2. Original List Final List
--------------------------------------------------------------
[12, 7] [13, 8]
[2, 3, 4, 5, 6] [3, 5, 5, 5, 6]
[3, 4, 5, 7, 9] [3, 5, 7, 9, 10]
[2, 3, 5, 7, 9] [3, 5, 7, 9, 10]
[4, 5, 9, 6, 2] [5, 7, 11, 7, 2]
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 void showSums(Scanner input) {
int count = 0;
int sum = 0;
while (input.hasNextInt()) {
sum += input.nextInt();
count++;
System.out.println("Sum of " + count + " = " + sum);
}
double average = (double) sum / count;
System.out.println("Average = " + average);
}
5. Line-Based File Processing. One possible solution appears below.
public static void underline(Scanner input) {
while (input.hasNextLine()) {
String text = input.nextLine();
if (!text.startsWith(".")) {
System.out.println(text);
} else {
System.out.println(text.substring(1));
for (int i = 1; i <= text.length() - 1; i++) {
System.out.print("-");
}
System.out.println();
}
}
}
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. Two possible solutions appear below.
public static void switchPairs(ArrayList<String> list) {
for (int i = 0; i < list.size() - 1; i += 2) {
String first = list.remove(i);
list.add(i + 1, first);
}
}
public static void switchPairs(ArrayList<String> list) {
int i = 0;
while (i < list.size() - 1) {
String first = list.get(i);
list.set(i, list.get(i + 1));
list.set(i + 1, first);
i += 2;
}
}
8. Critters. One possible solution appears below.
public class Chameleon extends Critter {
private int count = 0;
public Action getMove(CritterInfo info) {
count++;
if (count % 3 == 1) {
return Action.INFECT;
} else if (info.getFront() == Neighbor.EMPTY) {
return Action.HOP;
} else {
return Action.RIGHT;
}
}
public Color getColor() {
if (count % 3 == 0) {
return Color.RED;
} else if (count % 3 == 1) {
return Color.WHITE;
} else {
return Color.BLUE;
}
}
public String toString() {
if (count % 3 == 0) {
return "R";
} else if (count % 3 == 1) {
return "W";
} else {
return "B";
}
}
}
9. Arrays. One possible solution appears below.
public static int[] doubleSize(int[] list) {
int[] result = new int[2 * list.length];
for (int i = 0; i < list.length; i++) {
result[2 * i] = list[i] / 2 + list[i] % 2;
result[2 * i + 1] = list[i] / 2;
}
return result;
}
10. Programming. One possible solution appears below.
public static boolean hasTwoPair(int[] data) {
int[] counts = new int[6];
for (int i = 0; i < data.length; i++) {
counts[data[i] - 1]++;
}
int num = 0;
for (int i = 0; i < counts.length; i++) {
if (counts[i] == 2) {
num++;
}
}
return num == 2;
}
]]>
Stuart Reges
Last modified: Fri Sep 16 16:10:42 PDT 2011