Data Structure: Implementing Bubble Sort using C/C++
In Bubble sort, each pass consists of comparison each element in the file with its successor (i.e. x[i] with x[i+1]) and interchanging two elements if they are not in the proper order.
Example: Let us consider following array of elements
| 52 | 42 | 35 | 8 |
Following comparison are make on the first pass
| x[0] with x[1] (16 with 52) No interchange |
| x[1] with x[2] (52 with 42) Interchange |
| x[2] with x[3] (52 with 35) Interchange |
Thus after first pass, we can get: 16 42 35 8 52
- Note that after first pass, largest element (i.e. 52) get its proper position
- In general, x[n-1] will be in its proper position after iteration 1
We can list completer iterations as follows:
| Iteration 0: 16 52 42 35 8 |
| Iteration 1: 16 42 35 8 52 |
| Iteration 2: 16 35 8 42 52 |
| Iteration 3: 16 8 35 42 52 |
| Iteration 4: 8 16 35 42 52 |
Hence, for ith iteration, n-i iteration is required.
Algorithm
2. For i = 0 to i < (n – i), repeat step 3 to 4
3. For j = 0 to j < (n-i-1), repeat step 4
4. If x[j] > x[j+1] then swap the element as
temp = x[j]
x[j] = x[j+1]
x[j+1] = temp
5. Display the sorted array
Source code for Bubble Sort
#include<iostream>
using namespace std;
class BubbleSort{
private:
int no_of_elements;
int elements[10];
public:
void getarray();
void sortit();
void display();
};
void BubbleSort::getarray(){
cout<<"How many elements?: ";
cin>>no_of_elements;
cout<<"Insert array of element to sort: ";
for(int i=0;i<no_of_elements;i++){
cin>>elements[i];
}
}
void BubbleSort::sortit(){
int temp;
for(int i = 0; i < no_of_elements; i++){
for(int j =0; j < no_of_elements - 1 - i; j++){
if(elements[j] > elements[j+1]){
temp = elements[j];
elements[j] = elements[j+1];
elements[j+1] = temp;
}
}
}
}
void BubbleSort::display(){
cout<<"The sorted element is\n";
for(int i = 0 ; i < no_of_elements; i++){
cout<<elements[i]<<" ";
}
}
int main(){
BubbleSort BS;
BS.getarray();
BS.sortit();
BS.display();
return 0;
}
Efficiency of Bubble Sort:
The number of comparison between n elements is equal to (n-1) and total number of passes is also (n-1), The total number of comparison are therefore (n-1) * (n-1). Hence the efficiency of bubble sort is O(n^2)

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very helpful