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Solution to CSE142 Sample Final handout #31
1. Expression Value
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12/5 + 8/4 4
2.5 * 2 + 17/4 9.0
41 % 15 % 7 + 17 % 3 6
21/2 + "7 % 3" + 17 % 4 "107 % 31"
46/3/2.0/3 * 4/5 2.0
2. Original List Final List
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() ()
(7) (0)
(3, 2) (0, 2)
(5, 4, 3) (0, 4, 6)
(2, 4, 6, 8) (0, 4, 12, 24)
3. Polymorphism. The output produced is as follows.
b
c 1
a 2
b
c 1
b 2
c
c 1
c 2
b
d 1
b 2
4. Token-Based File Processing. One possible solution appears below.
public static void printStrings(Scanner input) {
while (input.hasNextInt()) {
int times = input.nextInt();
String word = input.next();
for (int i = 0; i < times; i++) {
System.out.print(word);
}
System.out.println();
}
}
5. Line-Based File Processing. One possible solution appears below.
public static void reverseLines(Scanner input) {
while (input.hasNextLine()) {
String text = input.nextLine();
for (int i = text.length() - 1; i >= 0; i--) {
System.out.print(text.charAt(i));
}
System.out.println();
}
}
6. Arrays. One possible solution appears below.
public static boolean isAllEven(int[] list) {
for (int i = 0; i < list.length; i++) {
if (list[i] % 2 != 0) {
return false;
}
}
return true;
}
7. ArrayLists. One possible solution appears below.
public static void removeShorterStrings(ArrayList<String> list) {
for (int i = 0; i < list.size() - 1; i++) {
String first = list.get(i);
String second = list.get(i + 1);
if (first.length() <= second.length()) {
list.remove(i);
} else {
list.remove(i + 1);
}
}
}
8. Critters. One possible solution appears below.
public class Pigeon implements Critter {
private int count;
private int max;
public char getChar() {
return 'P';
}
public int getMove(int x, int y) {
if (this.count == 2 * this.max) {
this.count = 0;
double n = Math.random();
if (n < 0.25) {
this.max = 2;
} else if (n < 0.5) {
this.max = 4;
} else if (n < 0.75) {
this.max = 6;
} else { // n > 0.75
this.max = 8;
}
}
this.count++;
if (this.count <= this.max) {
return SOUTH;
} else {
return NORTH;
}
}
}
9. Arrays. One possible solution appears below.
public static boolean isUnique(int[] list) {
for (int i = 0; i < list.length; i++) {
for (int j = i + 1; j < list.length; j++) {
if (list[i] == list[j]) {
return false;
}
}
}
return true;
}
10. Programming. One possible solution appears below.
public static int[] collapse(int[] list) {
int[] result = new int[list.length / 2 + list.length % 2];
for (int i = 0; i < list.length; i++) {
result[i / 2] += list[i];
}
return result;
}
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