I was trying another approach of sorting a linked list. Aside from the available methods, I've decided to take each node from the linked list and place it into an array. With that, I would be able to compare the data variables easily. I applied quickSort() on the array, and this is what I have. I would appreciate any feedback/comments on my code.

    public static void main(String[] args) {

        MyLinkedList list = new MyLinkedList();

        Student s = new Student(1, "John", 20, "Italy", "2011");
        Student s2 = new Student(2, "Mark", 19, "UAE", "2010");
        Student s3 = new Student(3, "Sally", 35, "UAE", "2000");

        System.out.println("Students in the list: ");

        Node[] n = list.convertA(list);
        quickSort(n, 0, (n.length-1));
        System.out.println("Sorted list is:");

        for(int q =0;q<n.length;q++){
            System.out.println(n[q] + " ");

    public static int partition(Node arr[], int left, int right) {

        int i = left, j = right;

        Node tmp;

        Node pivot = arr[(left + right) / 2];

        while (i <= j) {

            while (arr[i].getStudent().getAge() < pivot.getStudent().getAge()) {

            while (arr[j].getStudent().getAge() > pivot.getStudent().getAge()) {

            if (i <= j) {

                tmp = arr[i];
                arr[i] = arr[j];
                arr[j] = tmp;


        return i;


    public static void quickSort(Node arr[], int left, int right) {

        int index = partition(arr, left, right-1);

        if (left < index - 1) {
            quickSort(arr, left, index - 1);

        if (index < right) {
            quickSort(arr, index, right);

1 Answer 1


First of all I don't know why are you trying to reinvent the wheel? Java has a strong and beautiful API with many methods and there is also a method for sorting an custom array. Which is more faster than quicksort. However I think this is for your own practice.

  • Your convertA method is not good, cause you are passing the reference as argument as well. So either make the method static or use the existing method toArray().
  • You are converting a student list to an Node array, why? change Node array to Student array. So this


    will change to

  • there is a possible bug you are calculating (left + right) / 2 which may cause an overflow, though Java int range is very big you will face the overflow very rarely, but if you want to fix this; change to (right - left) / 2 + left.

Check How Other Have Done The Quicksort :

From CodeReview 1

From CodeReview 2

Why you should use the API method?


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