Why is my bubble sort algorithm implementation sorting the entire array and skips the first index?

Question

public static int[] bubbleSort(int[] inputArray){

        for(int i = 0; i < inputArray.length  - 1; i++ ){

            int tempa = inputArray[i];
            int tempb = inputArray[i + 1];

            if(inputArray[i] > inputArray[i + 1]){
                inputArray[i] = tempb;
                inputArray[i + 1] = tempa;
                i = 0;
                System.out.println(Arrays.toString(inputArray));
            }
        }
        return inputArray;
}

This implementation takes [20, 35, -15, 7, 55, 1, -22] and returns [20, -22, -15, 1, 7, 35, 55]. Sorts everything but the first index.

I'm new to Java. What I see here is that in worst case scenario the loop will restart for each element, hence n squared. Isn't that a bubble sort? — skypen
– skypen, Commented Sep 17, 2019 at 17:04

Andreas · Accepted Answer · 2019-09-17 17:08:50Z

1

Why ... skips the first index?

Because you set i = 0 inside the loop, but then the loop will do i++, so the first element is only examined on the first iteration, not on any "restart".

To restart correctly, use i = -1 so the i++ will make restart happen at i = 0, not at i = 1.

That will make the code work, but it is inefficient to restart immediately upon swapping two elements, because you will repeatedly recheck the beginning of the array.

answered Sep 17, 2019 at 17:08

Andreas

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2 Comments

skypen Over a year ago

Isn't that the definition of bubble sort? It's inefficient and and it checks the array n squared times. Do you have any other implementation of bubble sort so I could compare.

Andreas Over a year ago

I thought I had provided a link in a comment to the question, but I guess somebody deleted that. Try a web search for bubble sort, and look at e.g. the Wikipedia article. or the GeeksforGeeks article.

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