Bubble Sort

Algorithm :
Starting from the beginning of the list, compare every adjacent pair, swap their position if they are not in the right order (the latter one is smaller than the former one). After each iteration, one less element (the last one) is needed to be compared until there are no more elements left to be compared.

After first pass, largest element will be at the end of the array.

Ex:
Let us take the array of numbers "5 1 4 2 8", and sort the array from lowest number to greatest number using bubble sort algorithm. In each step, elements written in bold are being compared. Three passes will be required.
First Pass:
( 5 1 4 2 8 )  ( 1 5 4 2 8 ), Here, algorithm compares the first two elements, and swaps them.
( 1 5 4 2 8 )  ( 1 4 5 2 8 ), Swap since 5 > 4
( 1 4 5 2 8 )  ( 1 4 2 5 8 ), Swap since 5 > 2
( 1 4 2 5 8 )  ( 1 4 2 5 8 ), Now, since these elements are already in order (8 > 5), algorithm does not swap them.

Second Pass:
( 1 4 2 5 8 )  ( 1 4 2 5 8 )
( 1 4 2 5 8 )  ( 1 2 4 5 8 ), Swap since 4 > 2
( 1 2 4 5 8 )  ( 1 2 4 5 8 )
( 1 2 4 5 8 )  ( 1 2 4 5 8 )
Now, the array is already sorted, but our algorithm does not know if it is completed. The algorithm needs one whole pass without any swap to know it is sorted.

Third Pass:
( 1 2 4 5 8 )  ( 1 2 4 5 8 )
( 1 2 4 5 8 )  ( 1 2 4 5 8 )
( 1 2 4 5 8 )  ( 1 2 4 5 8 )
( 1 2 4 5 8 )  ( 1 2 4 5 8 )

Implementation:

class BubbleSort{
//method for the bubble sort
public static void bubbleSort(int a[], int size){
int temp = 0;
for(int i = 0; i < size-1; i++){
for(int j = 0; j < size-i-1; j++){
if(a[j] > a[j+1]){
temp = a[j];
a[j] = a[j+1];
a[j+1] = temp;
}
}
}
}

public static void main(String []args){
int a[] = {5,1, 4, 2, 8};

bubbleSort(a, a.length);
System.out.println("elements of array after sorting..."+java.util.Arrays.toString(a));
}
}