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Solution to CSE142 Final handout #32
1. Expression Value
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43 % 15/3 + 15/2 11
6.2 * 5/10 + 3.5 6.6
6 * 2.5/4 + (2.3 + 2.7)/4 5.0
"18" + 3 * 4 + (8 + 5) "181213"
59 % 10/(2 + 2) * 2.5/2 2.5
2. Original List Final List
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(5) (5)
(4, 7) (11, 7)
(2, 3, 4) (5, 7, 4)
(2, 4, 6, 8) (6, 10, 14, 8)
(9, 7, 5, 3, 1) (16, 12, 8, 4, 1)
3. Polymorphism. The output produced is as follows.
d
b 1
b 2
a
b 1
b 2
d
d 1
d 2
d
c 1
d 2
4. Token-Based File Processing. One possible solution appears below.
public static void reportRunningSum(Scanner input) {
double sum = input.nextDouble();
double max = sum;
System.out.print("running sum = " + sum);
while (input.hasNextDouble()) {
sum += input.nextDouble();
System.out.print(" " + sum);
if (sum > max) {
max = sum;
}
}
System.out.println();
System.out.println("max sum = " + max);
}
5. Line-Based File Processing. One possible solution appears below.
public static void listBlankLines(Scanner input) {
int line = 0;
int count = 0;
while (input.hasNextLine()) {
String text = input.nextLine();
line++;
if (text.length() == 0) {
System.out.println("line " + line + " is blank");
count++;
}
}
System.out.println("total blank lines = " + count);
}
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 Yak implements Critter {
private int move;
private int direction;
private int max;
public char getChar() {
return 'Y';
}
public int getMove(int x, int y) {
if (this.move == this.max) {
this.move = 0;
this.max++;
double n = Math.random();
if (n < 0.25) {
this.direction = NORTH;
} else if (n < 0.5) {
this.direction = SOUTH;
} else if (n < 0.75) {
this.direction = EAST;
} else { // n > 0.75
this.direction = WEST;
}
}
this.move++;
return this.direction;
}
}
9. Arrays. 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 removeZeros2(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;
}
}
The following solution would receive 10 out of 15 points.
public static void removeZeros(int[] list) {
int[] temp = new int[list.length];
int count = 0;
for (int i = 0; i < list.length; i++) {
if (list[i] != 0) {
temp[count] = list[i];
count++;
}
}
for (int i = 0; i < list.length; i++) {
list[i] = temp[i];
}
}
10. Programming. Two possible solutions appear below.
public static void rearrange(int[] list) {
int count = 0;
for (int i = 0; i < list.length; i++) {
if (list[i] % 3 == 0) {
swap(list, count, i);
count++;
}
}
for (int i = count; i < list.length; i++) {
if (list[i] % 3 == 1) {
swap(list, count, i);
count++;
}
}
}
public static void rearrange(int[] list) {
int index1 = 0;
int index2 = 0;
int index3 = list.length - 1;
while (index2 <= index3) {
if (list[index2] % 3 == 1) {
index2++;
} else if (list[index2] % 3 == 0) {
swap(list, index1, index2);
index1++;
index2++;
} else if (list[index3] % 3 == 2) {
index3--;
} else {
swap(list, index2, index3);
index3--;
}
}
}
Each of these solutions relies on the following "helper" method.
public static void swap(int[] list, int index1, int index2) {
int temp = list[index1];
list[index1] = list[index2];
list[index2] = temp;
}
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