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Sorting Algorithms |
Lecture Notes |
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Tech talks:
All tech talks can be found here
| TT0 | TT1 | TT2 | More Coming Soon! |
TT0:
challenge 1:
try catch block
code runs in try section, when an error occurs, catch block activates
in this case, we create an error in order for the catch section to activiate in case a different selection occurs
Use Exception e for catch all errors.
try catch block put in while loop to make sure correct input ensured unless user exits
try {
//wait time for loop, allow reader to read prompts and results
TimeUnit.SECONDS.sleep(1);
scan = new Scanner(System.in);
//prompt:
System.out.println("\nMenu page: ");
for(String i:options){
System.out.println("------------");
System.out.println(i);
}
System.out.println("Choose a program: ");
selection = scan.nextLine();
//checking selections
//quit programs
if (selection.equals( "0")) {
scan.close();
break;
}
//intswap selection
else if (selection.equals( "1")) {
System.out.println("Enter an integer: ");
int a = scan.nextInt();
System.out.println("Enter an integer: ");
int b = scan.nextInt();
IntByReference.swapper(a,b);
TimeUnit.SECONDS.sleep(3);
}
//array printer selection
else if (selection.equals( "2")) {
Matrix m0 = new Matrix(Matrix.keypad());
System.out.println("Keypad:");
System.out.println(m0);
Matrix m1 = new Matrix(Matrix.numbers());
System.out.println("Numbers Systems:");
System.out.println(m1);
TimeUnit.SECONDS.sleep(3);
}
//create an error for try catch block to activate and be useful
else {
String myString = null;
System.out.println(myString.length());
// System.out.println(options.length+1);
}
} catch(Exception e){
System.out.println("Incorrect input");
}
Challenge 2:
pass by value and pass by reference, had an extra variable to allow swapping, allowed user input for this program
had a constructor to hold values
use toString method to make sure the proper string is printed out.
challenge was pretty simple overall
public class IntByReference {
private int value;
public IntByReference(int a){
value =a;
}
public void swap(IntByReference a){
int value = this.value;
this.value=a.value;
a.value=value;
}
public String toString(){
return Integer.toString(value);
}
// Hack: create IntByReference, swapToLowHighOrder and toString methods
// static method that enables me to see numbers swapped by reference (before, after)
public static void swapper(int n0, int n1) {
IntByReference a = new IntByReference(n0);
IntByReference b = new IntByReference(n1);
System.out.println("Before: " + a + " " + b);
System.out.println("Swapped!");
a.swap(b); // conditionally build swap method to change values of a, b
System.out.println("After: " + a + " " + b);
System.out.println();
}
}
Challenge 3:
Created a constructor
Used nested for loops to traverse 2D arrays, one nest loop for traversing forward, one nested loop for traversing backwards
nest for loops in toString method to print and format data inside object
figured out when and where to use print new line and when not to
public String toString(){
//traverse matrix
for(int i = 0; i<matrix.length; i++){
for(int j =0; j<matrix[i].length; j++){
if(matrix[i][j]>=0) {
System.out.print(Integer.toHexString(matrix[i][j]));
System.out.print(" ");
}
else{
System.out.print(" ");
}
}
System.out.println(" ");
}
System.out.println("");
//traverse matrix from other side
for (int i =matrix.length-1; i>=0; i--){
for(int j=matrix[i].length-1; j>=0; j--){
if(matrix[i][j]>=0) {
System.out.print(Integer.toHexString(matrix[i][j]));
System.out.print(" ");
}
else{
System.out.print(" ");
}
}
System.out.println(" ");
}
return "";
}
TT1:
Challenge 1:
looked like a lot at first but was really simple
create delete method for forumula to delete elements in queue, took me a while to make sense of how queue works
created deletelist method to actually do the deleting, used nested for loops like in the addList
I find it weird how queues can't really be traversed like arrays
had to fiddle around with printqueue placement to get exact output
//other code not shown
public void addList(T[]... seriesOfObjects) {
for (T[] objects: seriesOfObjects)
for (T data : objects) {
System.out.println("Enqueued data: "+ data);
this.queue.add(data);
this.count++;
printQueue();
}
}
//other code not shown
public void deleteList(T[]... seriesOfObjects){
for(T[] objects: seriesOfObjects){
for(T data: objects){
System.out.println("Dequeued data: " + data);
this.queue.delete();
this.count--;
printQueue();
}
}
}
//other code not shown
public void delete(){
this.head=this.head.getNext();
if(head==null) {
LinkedList<T> tail = new LinkedList<>(null, null);
this.head=tail;
}
}
//other code not shown
Challenge 2:
This one was a bit difficult to figure out due to the fact that I couldn't traverse queue like an array
It also took me a while to figure out the properties of .poll(), didn't know it removed the head
When I figured it out, it was relatively easy and straightforward thinking, especially since the queue's themselves were already sorted
It was just a while statement comparing each head of the queue, then adding sorted values to a new queue
//other code not shown
Queue<Integer> mergedQueue = new LinkedList<Integer>();
while(!first.isEmpty() && !second.isEmpty()){
if(second.peek()>first.peek()){
mergedQueue.add(first.poll());
mergedQueue.add(second.poll());
}
else{
mergedQueue.add(second.poll());
mergedQueue.add(first.poll());
}
}
//other code not shown
Challenge 3
had to learn what a stack is, basically a type of data structure
basically just transfer queue items to stack and back to queue
just used for loops to do so, using size of queue as limit
//other code not shown
int n=a.size();
Stack<Integer> stack = new Stack<Integer>();
for(int i=0; i<n; i++){
int curr = a.poll();
stack.push(curr);
}
for (int i = 0; i < n; i++) {
int curr = stack.pop();
a.add(curr);
}
System.out.print("Reversed Queue: ");
for (Integer i : a) {
System.out.print(i + " ");
}
System.out.println();
//other code not shown
TT2:
Challenge 1:
Took a while for me to understand what was going on
basically code formats everything to reverse polish notation so computer can compile math, specifics will take while to explain
created an int checker to make sure token is number before pushing to stack
repeatable if else statements that corresponds to math expression
Operators Hashmap help figure out order of operations
happy this code wasn't done by scratch
public boolean intchecker(String n){
try{
double a = Double.parseDouble(n);
}catch(NumberFormatException e){
return false;
}
return true;
}
private void rpnToResult()
{
// Stack used to hold calculation while process RPN
Stack calculation = new Stack();
for(int i=0; i< reverse_polish.size(); i++)
{
if(intchecker(reverse_polish.get(i)))// If the token is a number
{
// Push number to stack
calculation.push(reverse_polish.get(i));
}
// else
else
{
// Pop the two top entries
Double fin = Double.parseDouble(calculation.pop().toString());
// Based off of Token operator calculate result
if(reverse_polish.get(i).equals("+")){
Double fin2 = Double.parseDouble(calculation.pop().toString());
result= fin + fin2;
}
else if(reverse_polish.get(i).equals("-")){
Double fin2 = Double.parseDouble(calculation.pop().toString());
result=fin-fin2;
}
else if(reverse_polish.get(i).equals("*")){
Double fin2 = Double.parseDouble(calculation.pop().toString());
result=fin*fin2;
}
else if(reverse_polish.get(i).equals("/")){
Double fin2 = Double.parseDouble(calculation.pop().toString());
result = fin/fin2;
}
else if(reverse_polish.get(i).equals("%")){
Double fin2 = Double.parseDouble(calculation.pop().toString());
result = fin%fin2;
}
else if(reverse_polish.get(i).equals("^")){
Double fin2 = Double.parseDouble(calculation.pop().toString());
result = Math.pow(fin2, fin);
}
else if(reverse_polish.get(i).equals("SQRT")){
result = Math.sqrt(fin);
}
// Push result back onto the stack
calculation.push(result);
}
}
// Pop final result and set as final result for expression
this.result=Double.parseDouble(calculation.pop().toString());
}
private final Map<String, Integer> OPERATORS = new HashMap<>();
{
// Map<"token", precedence>
OPERATORS.put("*", 3);
OPERATORS.put("/", 3);
OPERATORS.put("%", 3);
OPERATORS.put("+", 4);
OPERATORS.put("-", 4);
OPERATORS.put("^",2);
OPERATORS.put("SQRT", 2);
}
Challenge 2:
Doing exponent was easy
I create an expression just like the other operators
Used the math.pow util to help out, i aint about to write a whole algorithm
this was easy, my code can be seen in the snippet above
Extra Credit 1:
SQRT function was the same thing as the exponent function
just have to watch out for syntax if i ever have to do user input