# Bubble Sort Algorithm

The Bubble Sort algorithm is one the simplest sorting algorithm to implement. It works by continually looping through pairs in a list and comparing each pair. At each pair, it will perform a swap if the pair happens to be in the wrong order. The order will either be ascending or descending depending on what the programmer wants. This happens until no more swaps are needed. This algorithm apparently gets the name “bubble-sort” because you can imagine the smaller elements bubble to the top of the list. Checkout my bubble sort implementation below. When ran, it will give a good visual reference for the algorithm. It sorts in ascending order, complexity O(n^2). This algorithm approaches O(n) if the elements are already almost in order. Note that is the not the most efficient algorithm for large data sets.

```#include <iostream>

using namespace std;

int main()
{
int array[7];
int i, j;

for(i = 0; i <= 6; i++){
cout << "Enter unsorted numbers: " << endl;
cin >> array[i];
}
for(i = 0; i <= 5; i++){ //loop through pair of integers
for(j = i+1; j <= 6; j++){
int temp;
if(array[i] > array[j])
{
temp = array[i]; //swapping mechanisms
array[i] = array[j];
array[j] = temp;
}
}
}
for(i = 0; i <= 6; i++){
cout << array[i]; //sorted output
}
return 0;
}

```

# Fizz Buzz C++

This is a “classic” fizz-buzz solution I created. For those of you who don’t already know what fizz-buzz is, it follows this basic context: “Write a program that prints the numbers from 1 to 100. But for multiples of three print “Fizz” instead of the number and for the multiples of five print “Buzz”. For numbers which are multiples of both three and five print “FizzBuzz””. This question has been asked by recruiters as a method for determining programming competency.  While the fizz-buzz question is not really difficult itself, it does force you to think about one of the most important concepts to programmers, loop structures.

```
#include <iostream>

using namespace std;

int main(){
for (int i = 0; i <= 100; i++) {
if ((i % 15) == 0) //for multiples of both 3 and 5
cout << "FizzBuzz" << endl;
else if ((i % 3) == 0) //multiples of 3
cout << "Fizz" << endl;
else if ((i % 5) == 0) //multiples of 5
cout << "Buzz" << endl;
else
cout << i << endl; //all other numbers
}
return 0;
}

```

# C++ Primer Plus Chapter 7 Exercise 7

This exercise makes use heavier use of pointers then what we have been doing. However, much of this program can be pulled from listing 7.7 and around that part of the chapter. See source below.

Redo Listing 7.7, modifying the three array-handling functions to each use two pointer parameters to represent a range. The fill_array() function, instead of returning the actual number of items read, should return a pointer to the location after the last location filled; the other functions can use this pointer as the second argument to identify the end of the data.

```#include <iostream>

using namespace std;

const int Max = 5;

// function prototypes
double * fill_array(double *first, double *last);
void show_array(const double *first, const double *last);
void revalue(double *first, double *last, double factor);

int main()
{
double properties[Max];
double * last;

last = fill_array(properties, properties+Max);
show_array(properties, last);

cout << "Enter revaluation factor: ";
double factor;
cin >> factor;

revalue(properties, last, factor);
show_array(properties, last);
cout << "Done.\n";

return 0;
}

double *fill_array(double *first, double *last)
{
double temp;
double *pt;
int i=0;
for (pt = first; pt != last; pt++, i++)
{
cout << "Enter value #" << (i + 1) << ": ";
cin >> temp;
{
cin.clear();
while (cin.get() != '\n')
continue;
cout << "Bad input; input process terminated.\n";
break;
}
else if (temp < 0) // signal to terminate
break;
*pt = temp;
}
return pt;
}

void show_array(const double *first, const double *last)
{
const double *pt;
int i=0;
for (pt = first; pt != last; pt++, i++)
{
cout << "Property #" << (i + 1) << ": \$ \n";
cout << *pt;
}
}

void revalue(double *first, double *last, double factor)
{
double *pt;
for (pt = first; pt != last; pt++)
*pt *= factor;
}

```