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Key to CSE142 Final handout #22
Winter 2009
1. Expression Value
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8.0/4/4 + 3/4 + 0.5 * 3 2.0
88 % 10 % 5/2 + 7 % 8 8
2 * 3 + 4 + " = " + 5 * 6 + 7 "10 = 307"
54/5/2.0/4 * 3 3.75
2 * (0.75 + 1.25)/5 * 10 - 7/4 7.0
2. Original List Final List
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{6, 3} {7, 3}
{2, 4, 6} {3, 3, 6}
{1, 2, 3, 4} {2, 1, 4, 4}
{2, 2, 2, 2, 2} {3, 1, 3, 1, 2}
{7, 3, 2, 0, 5, 1} {8, 2, 3, -1, 6, 1}
3. Polymorphism. The output produced is as follows.
b
c 1
b 2
a
c 1
d 2
d
c 1
d 2
b
b 1
b 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 processScores(Scanner input) {
while (input.hasNextLine()) {
String name = input.nextLine();
String data = input.nextLine();
int plus = 0;
int count = 0;
for (int i = 0; i < data.length(); i++) {
count++;
if (data.charAt(i) == '+') {
plus++;
}
}
double percent = 100.0 * plus / count;
System.out.println(name + ": " + percent + "% plus");
}
}
6. Arrays. One possible solution appears below.
public static boolean isFibLike(int[] list) {
for (int i = 2; i < list.length; i++) {
if (list[i] != list[i - 2] + 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 Orca extends Critter {
int count;
public Orca() {
count = 0;
}
public Action getMove(CritterInfo info) {
if (count % 6 == 4 || count % 6 == 5) {
count++;
return Action.LEFT;
} else if (info.getFront() != Neighbor.EMPTY) {
return Action.INFECT;
} else {
count++;
return Action.HOP;
}
}
public String toString() {
if (count % 6 < 4) {
return "M";
} else {
return "T";
}
}
}
9. Arrays. One possible solution appears below.
public static int[] append(int[] list1, int[] list2) {
int[] result = new int[list1.length + list2.length];
for (int i = 0; i < list1.length; i++) {
result[i] = list1[i];
}
for (int j = 0; j < list2.length; j++) {
result[list1.length + j] = list2[j];
}
return result;
}
10. Programming. One possible solution appears below.
public static String acronym(String s) {
boolean inWord = false;
s = s.toUpperCase();
String result = "";
for (int i = 0; i < s.length(); i++) {
char ch = s.charAt(i);
if (ch == ' ' || ch == '-') {
inWord = false;
} else if (!inWord) {
inWord = true;
result += ch;
}
}
return result;
}
]]>
Stuart Reges
Last modified: Fri Sep 16 16:10:08 PDT 2011