Unit 4 - Iteration:

  • This assignement will be made of 3 multiple choice questions, and 2 coding hacks.

Question 1:

What does the following code print?

A. 5 6 7 8 9

B. 4 5 6 7 8 9 10 11 12

C. 3 5 7 9 11

D. 3 4 5 6 7 8 9 10 11 12

Answer: D

for (int i = 3; i <= 12; i++)
{
   System.out.print(i + " ");
}

3 4 5 6 7 8 9 10 11 12

Question 2:

How many times does the following method print a *?

A. 9

B. 7

C. 6

D. 10

Answer: C

for (int i = 3; i < 9; i++)
{
   System.out.print("*");
}

******

Question 3:

What does the following code print?

A. 5 4 3 2 1

B. -5 -4 -3 -2 -1

C. -4 -3 -2 -1 0

Answer: C

int x = -5;
while (x < 0)
{
   x++;
   System.out.print(x + " ");
}

-4 -3 -2 -1 0

Loops homework hack

Easy Hack

  • use a while loop to find the numbers from 1-50 that are divisible by 3 or 5, then store them into a list (make sure to print it out at the end)
  • use a for loop to do the same thing detailed above
import java.util.ArrayList;

ArrayList<Integer> arr = new ArrayList<>();
int i = 0;
while (i <= 50) {
    if (i%3==0 || i%5==0) arr.add(i);
    i++;
}
System.out.println(arr);

ArrayList<Integer> arr2 = new ArrayList<>();
for (int i = 0; i <= 50; i++) {
    if (i%3==0 || i%5==0) arr2.add(i);
}
System.out.println(arr2);

[0, 3, 5, 6, 9, 10, 12, 15, 18, 20, 21, 24, 25, 27, 30, 33, 35, 36, 39, 40, 42, 45, 48, 50]
[0, 3, 5, 6, 9, 10, 12, 15, 18, 20, 21, 24, 25, 27, 30, 33, 35, 36, 39, 40, 42, 45, 48, 50]

Harder Hack

Palindromes are numbers that have the same value when reversed (ex: “123321” or “323”). Create a program that uses a while loop that outputs all palindromes in any given list.

Sample Input: test_list = [5672, 235, 5537, 6032, 317, 8460, 1672, 8104, 7770, 4442, 913, 2508, 1116, 9969, 9091, 522, 8756, 9527, 7968, 1520, 4444, 515, 2882, 6556, 595]

Sample Output: 4444, 515, 2882, 6556, 595

public String reverse(String str) {
    String reversed = "";
    for (int i = str.length()-1; i >= 0; i--) {
        reversed += str.charAt(i);
    }
    return reversed;
}

int[] test_list = {5672, 235, 5537, 6032, 317, 8460, 1672, 8104, 7770, 4442, 913, 2508, 1116, 9969, 9091, 522, 8756, 9527, 7968, 1520, 4444, 515, 2882, 6556, 595};

System.out.println("Palindromes: ");
int i = 0;
while (i < test_list.length) {
    int num = test_list[i];
    String strNum = Integer.toString(num);
    if (strNum.equals(reverse(strNum))) {
        System.out.print(num + " ");
    }
    i++;
}

Palindromes: 
4444 515 2882 6556 595

Coding Hack(For above 0.9):

Use a for loop to output a spiral matrix with size n

For example: Sample Input: 3

Output: [[1, 2, 3], [8, 9, 4], [7, 6, 5]]

import java.util.ArrayList;

int n = 3;

ArrayList<ArrayList<Integer>> mtx = new ArrayList<>();
for (int i = 0; i < n; i++) {
    mtx.add(new ArrayList<Integer>());
    for (int j = 0; j < n; j++) {
        mtx.get(i).add(0);
    }
}

int index = 1;
int top = 0;
int bottom = n - 1;
int left = 0;
int right = n - 1;

while (index <= n*n) {
    for (int j = left; j <= right && index <= n * n; j++) {
        mtx.get(top).set(j, index++);
    }
    top++;

    for (int i = top; i <= bottom && index <= n * n; i++) {
        mtx.get(i).set(right, index++);
    }
    right--;

    for (int j = right; j >= left && index <= n * n; j--) {
        mtx.get(bottom).set(j, index++);
    }
    bottom--;

    for (int i = bottom; i >= top && index <= n * n; i--) {
        mtx.get(i).set(left, index++);
    }
    left++;
}

for (ArrayList<Integer> row : mtx) {
    for (Integer num : row) {
        System.out.print(num + " ");
    }
    System.out.println();
}


1 2 3 
8 9 4 
7 6 5