Review of linked list and quick sort

Question

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");
 list.addStudent(s);
 Student s2 = new Student(2, "Mark", 19, "UAE", "2010");
 list.addStudent(s2);
 Student s3 = new Student(3, "Sally", 35, "UAE", "2000");
 list.addStudent(s3);
 System.out.println("Students in the list: ");
 list.print();
 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()) {
 i++;
 }
 while (arr[j].getStudent().getAge() > pivot.getStudent().getAge()) {
 j--;
 }
 if (i <= j) {
 tmp = arr[i];
 arr[i] = arr[j];
 arr[j] = tmp;
 i++;
 j--;
 }
 }
 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);
 }
 }

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

  arr[i].getStudent().getAge()
  

  will change to

  arr[i].getAge() 
  

• 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?