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Solution to CSE142 Final handout #29
Spring 2006
1. Expression Value
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17/3 + 55 % 20/2 12
19.0/2 + 2 * 2.25 14.0
2 + 2 + "." + 2 + 2 + 4 * 5 "4.2220"
73/10 * 4/3/2.0 * 3 13.5
15/2 * (1.5 + 2.5) + 98 % 5/2 29.0
2. Original List Final List
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(8) (8)
(6, 3) (6, 9)
(2, 4, 6) (2, 6, 12)
(1, 2, 3, 4) (1, 3, 6, 10)
(7, 3, 2, 0, 5) (7, 10, 12, 12, 17)
3. Polymorphism. The output produced is as follows.
d
d 1
b 2
c
c 1
c 2
c
c 1
a 2
d
d 1
a 2
4. Token-Based File Processing. One possible solution appears below.
public static void processSets(Scanner input) {
int count = 0;
double sum = 0.0;
while (input.hasNextInt()) {
int n = input.nextInt();
if (n == 0) {
System.out.println("total values = " + count);
System.out.println("average value = " + sum / count);
System.out.println();
count = 0;
sum = 0;
} else {
count += n;
sum += n * input.nextDouble();
}
}
}
5. Line-Based File Processing. One possible solution appears below.
public static void countOddLength(Scanner input) {
int odd = 0;
int total = 0;
while (input.hasNextLine()) {
String text = input.nextLine();
total++;
if (text.length() % 2 != 0) {
odd++;
}
}
System.out.println("Total lines = " + total);
System.out.println("odd length = " + odd);
double percent = odd * 100.0 / total;
System.out.println("percent = " + percent);
}
6. Arrays. One possible solution appears below.
public static int numUnique(int[] list) {
if (list.length == 0) {
return 0;
} else {
int count = 1;
for (int i = 1; i < list.length; i++) {
if (list[i] != list[i - 1]) {
count++;
}
}
return count;
}
}
7. ArrayLists. One possible solution appears below.
public static void collapse(ArrayList<String> list) {
for (int i = 0; i < list.size() - 1; i++) {
String pair = "(" + list.get(i) + ", " + list.get(i + 1) + ")";
list.set(i, pair);
list.remove(i + 1);
}
}
8. Critters. One possible solution appears below.
public class Orca implements Critter {
private int move;
private int max;
private char display = '+';
public char getChar() {
return display;
}
public int getMove(CritterInfo info) {
if (this.move == 3 * this.max) {
this.move = 0;
this.max++;
}
this.move++;
if (this.move <= this.max) {
display = '+';
return CENTER;
} else if (this.move <= 2 * this.max) {
display = '<';
return WEST;
} else {
display = '>';
return EAST;
}
}
}
9. Arrays. One possible solution appears below.
public static void insert(int[] shortList, int[] bigList, int index) {
int lastIndex = bigList.length - 1 - shortList.length;
for (int i = lastIndex; i >= index; i--) {
bigList[i + shortList.length] = bigList[i];
}
for (int i = 0; i < shortList.length; i++) {
bigList[index + i] = shortList[i];
}
}
10. Programming. One possible solution appears below.
public static void printReversed(String s) {
boolean inWord = false;
int start = -1;
for (int i = 0; i < s.length(); i++) {
if (s.charAt(i) == ' ') {
inWord = false;
System.out.print(' ');
} else {
if (!inWord) {
inWord = true;
start = i;
}
if (i == s.length() - 1 || s.charAt(i + 1) == ' ') {
for (int j = i; j >= start; j--) {
System.out.print(s.charAt(j));
}
}
}
}
System.out.println();
}
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