How to understand Bubble Sort Pseudo code?

[Gnabster]

Hi,

I have difficulity understanding the Pseudo code compared to Java, i had cracked my head but still dont understand the coding, just a few lines but i couldn't get myself to fully understand.

The Pseudo Code:

for i=1 to n-1 do
for j=i to n-1 do
if x[j]>x[j+1] then
temp=x[j]
x[j]=x[j+1]
x[j+1]=temp
end {if}
end {for}
end {for}


From my understanding of this bubble sort code:

int[] x = {5, 4, 3, 2, 1};
int n = x.length;
int temp;
int counter = 0;

for (int i = 0; i < n; i++) {
// counter ++;
for (int j = i; j < n; j++) {

if (x[j] > x[j + 1]) {
temp = x[j];
x[j] = x[j + 1];
x[j + 1] = temp;
counter++;
}
}
}

for (int i = 0; i < x.length; i++) {
System.out.print(x[i] + " ");
}

System.out.println();
System.out.println(counter);


My question is:

Is the Pseudo Code same as the other bubble sort code online? This do not have the true/false checker.
I need to calculate the:

1) Best case
2) Worst case
3) Average case

for the pseudo code, in terms of Big O Notation, all bubble sort should be worst case Big O N square, best is Big O(n), average is same as worst case.
However i am unable to convince myself by looking at the pseudo code alone, any kind souls please assist if you understand the Pseudo code, Thanks in advance

[DarrylBurke]

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[JosAH]

for the pseudo code, in terms of Big O Notation, all bubble sort should be worst case Big O N square, best is Big O(n), average is same as worst case.
However i am unable to convince myself by looking at the pseudo code alone, any kind souls please assist if you understand the Pseudo code, Thanks in advance

Best, average and worst case are all O(n^2); if you're just counting the number of swaps the best case is O(0) (no swaps needed if the array is already sorted), but the average and worst case are still O(n^2).

kind regards,

Jos