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Key to CSE143X Sample Midterm, Fall 2012 handout #10
1. Expression Value
-----------------------------------------------
4 * (2 + 4) - 3 * 5 9
54 % 10 + 8 * 3 % 9 10
3 * 2 + 4 + "+" + 2 + 3 * 4 "10+212"
2.3 * 3 + 19 / 5 / 2 + 6.0 / 5 9.1
108 / 20 * 3 / 4 / 2.0 + 1.0 / 2 2.0
2. Parameter Mystery. The program produces the following output.
students in the beer = 12
beers in the dorm = 125
students in the dorm = 250
beers in the student = 4
3. Method Call Output Produced
---------------------------------------
ifElseMystery(4, 15); 15 18
ifElseMystery(7, 17); 27 10
ifElseMystery(12, 5); 15 10
ifElseMystery(16, 8); 28 2
4. Method Call Value Returned
--------------------------------------
mystery(6, 0) 0
mystery(8, 1) 8
mystery(3, 3) 17
mystery(4, 2) 10
mystery(1, 4) 24
5. next < 0 next == 0 x > 0
+---------------------+---------------------+---------------------+
Point A | sometimes | sometimes | never |
+---------------------+---------------------+---------------------+
Point B | sometimes | never | sometimes |
+---------------------+---------------------+---------------------+
Point C | always | never | always |
+---------------------+---------------------+---------------------+
Point D | sometimes | sometimes | sometimes |
+---------------------+---------------------+---------------------+
Point E | never | always | sometimes |
+---------------------+---------------------+---------------------+
6. Three possible solutions appear below.
public static boolean hasMidpoint(int x, int y, int z) {
return (2 * x == y + z || 2 * y == x + z || 2 * z == x + y);
}
public static boolean hasMidpoint(int x, int y, int z) {
return (x - y == y - z || x - z == z - y || y - x == x - z);
}
public static boolean hasMidpoint(int x, int y, int z) {
double average = (x + y + z) / 3.0;
return (x == average) || (y == average) || (z == average);
}
7. One possible solution appears below.
public static void flip(Random r, int n) {
System.out.print("flips:");
int count = 0;
while (count < n) {
int result = r.nextInt(2);
if (result == 0) {
System.out.print(" H");
count++;
} else {
System.out.print(" T");
count = 0;
}
}
System.out.println();
}
8. 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;
}
}
9. One possible solution appears below.
public static boolean samePattern(String s1, String s2) {
if (s1.length() != s2.length())
return false;
for (int i = 0; i < s1.length(); i++)
for (int j = i + 1; j < s2.length(); j++)
if ((s1.charAt(i) == s1.charAt(j)) !=
(s2.charAt(i) == s2.charAt(j)))
return false;
return true;
}
}
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