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Is it possible to change my recursive method to iterative?

I'm writing a quicksort algorithm to sort numbers with random pivot. How can I change my quicksort methods from recursive to iterative? I have one sort method which is recursive, but I need the iterative method. Is it possible to change from recursive to iterative in just the sort method or do I have to change the whole code?
here is all my code
public class Main {
public static void main(String[] args) {
long start = System.currentTimeMillis();
ArrayList time = new ArrayList();
for (int k = 1; k < 1000000; k++) {
time.add(k);
}
int[] tall = new int[1000000];
int index = 0;
int n = 1000000;
File text = new File("/Users/sasan/IdeaProjects/File.txt");
try {
Scanner scan = new Scanner(text);
while (scan.hasNextLine() && index < 1000000) {
tall[index] = scan.nextInt();
index++;
}
scan.close();
} catch (IOException e) {
System.out.println("Problem with file");
e.printStackTrace();
}
int l = tall.length;
sort(tall, 0, l-1);
System.out.println("Sorted array");
printArray(tall);
System.out.println("");
long end = System.currentTimeMillis();
System.out.print("Execution Time is: ");
System.out.print((end - start));
}
static void random(int tall[],int low,int high)
{
Random rand= new Random();
int pivot = rand.nextInt(high-low)+low;
int temp1=tall[pivot];
tall[pivot]=tall[high];
tall[high]=temp1;
}
/* This function takes last element as pivot,
places the pivot element at its correct
position in sorted array, and places all
smaller (smaller than pivot) to left of
pivot and all greater elements to right
of pivot */
static int partition(int tall[], int low, int high)
{
// pivot is choosen randomly
random(tall,low,high);
int pivot = tall[high];
int i = (low-1); // index of smaller element
for (int j = low; j < high; j++)
{
// If current element is smaller than or
// equal to pivot
if (tall[j] < pivot)
{
i++;
// swap arr[i] and arr[j]
int temp = tall[i];
tall[i] = tall[j];
tall[j] = temp;
}
}
// swap arr[i+1] and arr[high] (or pivot)
int temp = tall[i+1];
tall[i+1] = tall[high];
tall[high] = temp;
return i+1;
}
/* The main function that implements QuickSort()
tall[] --> Array to be sorted,
low --> Starting index,
high --> Ending index */
static void sort(int tall[], int low, int high)
{
if (low < high)
{
/* pi is partitioning index, tall[pi] is
now at right place */
int pi = partition(tall, low, high);
// Recursively sort elements before
// partition and after partition
sort(tall, low, pi-1);
sort(tall, pi+1, high);
}
}
/* A utility function to print array of size n */
static void printArray(int tall[])
{
int n = tall.length;
for (int i = 0; i < n; ++i)
System.out.print(tall[i]+" ");
System.out.println();
}
// Driver Code
}

You can generally always turn a recursive method into an iterative method, though the point about using recursion is that you usually choose it when it's most convenient to do so. A divide-and-conquer algorithm such as quicksort is precisely a good candidate for recursion.
To turn a recursive method into an iterative method, you basically use a 'stack' object (and Java provides the Stack class to assist with this): this essentially stores what parameters need to be operated on on each iteration. Then, whereas before you would have called into your recursive method, you put on top of your stack the parameters that would have been used to call it. Each iteration does the equivalent of "returning" from the method by taking the latest parameters off the top of the stack. (Essentially, you're simulating the process that Java goes through under the hood when you call a method.)
So the pattern looks like this:
Stack<Paraneters> stack = new Stack();
stack.push(new Parameters(0, size);
while (!stack.empty()) {
Parameters params = stack.pop();
int lo = params.lo;
int hi = params.hi;
if (lo < hi) {
int pivot = partition(lo, hi);
stack.push(new Parameters(lo, pivot-1));
stack.push(new Parameters(pivot+1, hi));
}
}
For this simple example, the Parameters object could actually just be a 2-length int array, but for more complex combinations of parameters you may end up constructing a wrapper class. (Another reason why using recursion can be more convenient in practice.)

kthSmallest method when an array/sub-array of integers is partitioned two parts with pivot at the end

So I think my partition method works but I cannot understand or figure out how to do the kthSmallest method. I no longer get out of bounds errors with my partition method which leads me to think that it works and with testing it seems to work. However, my kthSmallest method often gets stuck in an infinite loop and when it does return a value, it is never the correct value.
I have seen examples online that place the pivot between the two subarrays however for our assignment the pivot is always at the end so I often get confused looking at these examples.
Here is what I have:
class Median{
static int kthSmallest(int[] arr, int left, int right, int k){
int divider = partition(arr, left, right);
if(divider == k-1){
return arr[right];
}
else if(divider > k-1){
return kthSmallest(arr, left, divider-1, k);
}
else{
return kthSmallest(arr, divider, right, (k - divider-1));
}
}
static int partition(int[] arr, int left, int right){
int pivot = arr[right];
int index = left;
for(int i = left; i < right-1; i++){
if(arr[i] < pivot){
swap(arr, index, i);
index++;
}
}
printArr(arr);
System.out.println("divider: " + index);
return index;
}
static void swap(int[] arr, int i, int j){
int temp = arr[i];
arr[i] = arr[j];
arr[j] = temp;
}
static void printArr(int[] arr){
System.out.println();
for(int index = 0; index < arr.length - 1; index++){
System.out.print(arr[index] + ", ");
}
System.out.print(arr[arr.length-1]);
System.out.println();
}
public static void main(String[] args){
int[] arr = {34,-1,0,5,3};
printArr(arr);
//System.out.println(partition(arr, 0, arr.length-1));
//printArr(arr);
System.out.println(kthSmallest(arr, 0, arr.length - 1, 2));
}
}

It seems like what you are trying to do is implement the QuickSelect algorithm, so I recommend taking a look at the Wikipedia article.
Looking at your code, I think you have misinterpreted the "pivot at the end" bit. I believe that what your requirements want is to select the pivot from the end, and then place it in between the two sublists so that your list looks right. For example,
34 -1 0 5 3
should become
-1 0 3 34 5
not
-1 0 34 5 3, where your pivot is not doing its job correctly.
There are also a few problems with your kthSmallest method. Double check the values you pass along in your recursion and you also are missing a case which might cause infinite recursion. Spoilers for if you are absolutely stuck:
You since you don't re-index the list, you should change '(k - divider -1)' to just k.
If left == right then you get unnecessary recursion, so you should just return in that case.
In your partition method, make sure you iterate far enough in the part of the list you are partitioning. Again, just in case you're really stumped:
for(int i = left; i < right-1; i++) should become for(int i = left; i <= right-1; i++)

Inplace Quicksort in Java

For refreshing some Java I tried to implement a quicksort (inplace) algorithm that can sort integer arrays. Following is the code I've got so far. You can call it by sort(a,0,a.length-1).
This code obviously fails (gets into an infinite loop) if both 'pointers' i,j point each to an array entry that have the same values as the pivot. The pivot element v is always the right most of the current partition (the one with the greatest index).
But I just cannot figure out how to avoid that, does anyone see a solution?
static void sort(int a[], int left, int right) {
if (right > left){
int i=left, j=right-1, tmp;
int v = a[right]; //pivot
int counter = 0;
do {
while(a[i]<v)i++;
while(j>0 && a[j]>v)j--;
if( i < j){
tmp = a[i];
a[i] = a[j];
a[j] = tmp;
}
} while(i < j);
tmp = a[right];
a[right] = a[i];
a[i] = tmp;
sort(a,left,i-1);
sort(a,i+1,right);
}
}

When preforming a Quicksort I strongly suggest making a separate method for partitioning to make the code easier to follow (I'll show an example below). On top of this a good way of avoiding worst case run time is shuffling the array you're sorting prior to preforming the quick sort. Also I used the first index as the partitioning item instead of the last.
For example:
public static void sort (int[] a)
{
StdRandom.shuffle(a);
sort(a, 0, a.length - 1);
}
private static void sort(int[] a, int lo, int hi)
{
if (hi <= lo) return;
int j = partition(a, lo, hi) // the addition of a partitioning method
sort(a, lo, j-1);
sort(a, j+1, hi);
}
private static int partition(int[] a, int lo, int hi)
{
int i = lo, j = hi + 1, tmp = 0;
int v = a[lo];
while (true)
{
while (a[i++] < v) if (i == hi) break;
while (v < a[j--]) if (j == lo) break;
if (i >= j) break;
tmp = a[i];
a[i] = a[j];
a[j] = tmp;
}
tmp = a[lo];
a[lo] = a[j];
a[j] = temp;
return j;
}
On top of this if you want a really good example on how Quicksort works (as a refresher) see here.

This should work (will check for correctness in a bit, it works!):
EDIT: I previously made a mistake in error checking. I forgot to add 2 more conditions, here is the amended code.
public static void main (String[] args) throws java.lang.Exception
{
int b[] = {10, 9, 8, 7, 7, 7, 7, 3, 2, 1};
sort(b,0,b.length-1);
System.out.println(Arrays.toString(b));
}
static void sort(int a[], int left, int right) {
if (right > left){
int i=left, j=right, tmp;
//we want j to be right, not right-1 since that leaves out a number during recursion
int v = a[right]; //pivot
do {
while(a[i]<v)
i++;
while(a[j]>v)
//no need to check for 0, the right condition for recursion is the 2 if statements below.
j--;
if( i <= j){ //your code was i<j
tmp = a[i];
a[i] = a[j];
a[j] = tmp;
i++;
j--;
//we need to +/- both i,j, else it will stick at 0 or be same number
}
} while(i <= j); //your code was i<j, hence infinite loop on 0 case
//you had a swap here, I don't think it's needed.
//this is the 2 conditions we need to avoid infinite loops
// check if left < j, if it isn't, it's already sorted. Done
if(left < j) sort(a,left,j);
//check if i is less than right, if it isn't it's already sorted. Done
// here i is now the 'middle index', the slice for divide and conquer.
if(i < right) sort(a,i,right);
}
}
This Code in the IDEOne online compiler
Basically we make sure that we also swap the value if the value of i/j is the same as the pivot, and break out of the recursion.
Also there was a check in the pseudocode for the length, as if we have an array of just 1 item it's already sorted (we forgot the base case), I thought we needed that but since you pass in the indexes and the entire array, not the subarray, we just increment i and j so the algorithm won't stick at 0 (they're done sorting) but still keep sorting an array of 1. :)
Also, we had to add 2 conditions to check if the array is already sorted for the recursive calls. without it, we'll end up sorting an already sorted array forever, hence another infinite loop. see how I added checks for if left less than j and if i less than right. Also, at that point of passing in i and j, i is effectively the middle index we split for divide and conquer, and j would be the value right before the middle value.
The pseudocode for it is taken from RosettaCode:
function quicksort(array)
if length(array) > 1
pivot := select any element of array
left := first index of array
right := last index of array
while left ≤ right
while array[left] < pivot
left := left + 1
while array[right] > pivot
right := right - 1
if left ≤ right
swap array[left] with array[right]
left := left + 1
right := right - 1
quicksort(array from first index to right)
quicksort(array from left to last index)
Reference: This SO question
Also read this for a quick refresher, it's implemented differently with an oridnary while loop
This was fun :)

Heres some simple code I wrote that doesn't initialize to many pointers and gets the job done in a simple manner.
public int[] quickSort(int[] x ){
quickSortWorker(x,0,x.length-1);
return x;
}
private int[] quickSortWorker(int[] x, int lb, int ub){
if (lb>=ub) return x;
int pivotIndex = lb;
for (int i = lb+1 ; i<=ub; i++){
if (x[i]<=x[pivotIndex]){
swap(x,pivotIndex,i);
swap(x,i,pivotIndex+1);
pivotIndex++;
}
}
quickSortWorker(x,lb,pivotIndex-1);
quickSortWorker(x,pivotIndex+1,ub);
return x;
}
private void swap(int[] x,int a, int b){
int tmp = x[a];
x[a]=x[b];
x[b]=tmp;
}

Quicksort - recursive

I am trying to code quicksort in two ways, one in-place, and the other by using separate arrays. I'm kind of stuck on some of the logic, take a look at what I have, Thanks for the help in advance!
public List<Integer> sort(List<Integer> arr){
if(arr.length > 0)
List<Integer> ret = new ArrayList<Integer>();
ret = quickSort(arr);
return ret;
}
public List<Integer> quickSort(List<Integer> arr){
if(arr.length < 2)
return;
int pivot = arr[0];
List<Integer> left = new ArrayList<Integer>();
List<Integer> right = new ArrayList<Integer>();
for(int i = 0; i < arr.length; i++){
if(arr[i] <= pivot)
left.add(arr[i]);
else
right.add(arr[i]);
}
quickSort(left);
quickSort(right);
}
Now i'm stuck, I don't know what I would do after recursively going through both sets, mostly stuck on how would I connect them together and return a sorted list.

You need to combine left and right sequences together. You need to do it at the end of your algorithm (before the closing }). In pseudo code:
int leftpos = 0, rightpos = 0;
List newlist = new ArrayList();
for(int pos = 0; pos < arr.length; pos++)
if left[pos] < right[pos] newlist.add(left[leftpos++]);
else newlist.add(right[rightpos++]);
return newlist;
This is just a pseudo-code. You need to add code to check lengths of each array (left and right) in the for cycle.
Also I must note that this is far from quicksort. So many new array allocations make the algorithm extremely slow and that's unwelcome when sorting.
Also, right side of line 3 is redundant. You don't need to allocate anything here, as it is overwritten in the next line. I would just simply replace your lines 3-5 with this:
return quickSort(arr);

Let me have a crack at this for you.
First off, you always want to do in-place sorting unless you're working with linked lists (and even then it usually pays to convert to an array, sort in place, then convert back to a linked list -- it puts way less pressure on the garbage collector). .NET List<>s are actually expanding arrays.
Next, quicksort is really all about the pivot operation. Here's one way to do it:
// Quicksort the sub-array xs[lo..hi].
void QSort(int[] xs, int lo, int hi) {
if (hi <= lo) return; // Don't sort empty or singleton sub-arrays.
var p = [choose some pivot value from xs[lo..hi]];
var a = lo; // Invariant: x[lo..a - 1] <= p.
var z = hi; // Invariant: p < x[z + 1..hi].
while (a <= z) {
if (xs[a] <= p) a++; else Swap(xs, a, z--);
}
QSort(xs, lo, a - 1); // Sort the items <= p.
QSort(xs, z + 1, hi); // Sort the items > p.
}
void Swap(int[] xs, int i, int j) {
var tmp = xs[i];
xs[i] = xs[j];
xs[j] = tmp;
}

Simple implementation on Groovy
def qs(list) {
if (list.size() < 2) return list
def pivot = list[0]
def items = list.groupBy { it <=> pivot }.withDefault { [] }
qs(items[-1]) + items[0] + qs(items[1])
}

Got stackoverflowerror when using quickSort, can I increase the stack and the heap?

Can I increase the stack and the heap in java? I'm using BlueJ.
========
EDIT:
Here is the code:
// ***** Quick-Sort Method *****
public static void quickSort(int[] data, int first, int n)
{
int p, n1, n2;
if(n > 1)
{
p = partition(data, first, n);
n1 = p - first;
n2 = n - n1 - 1;
quickSort(data, first, n1);
quickSort(data, p+1, n2);
}
}
// ***** PRIVATE HELPER FUNCTIONS *****
public static void quickSort(int[] data)
{
quickSort(data, 0, data.length);
}
private static int partition(int[] A, int first, int n )
{
int right = first + n - 1;
int ls = first;
int pivot = A[first];
for(int i = first+1; i <= right; i++)
{
if(A[i] <= pivot)
// Move items smaller than pivot only, to location that would be at left of pivot
{
ls++;
swap(A, i, ls);
}
}
swap(A, first, ls);
return ls;
}
private static void swap(int[] data, int pos1, int pos2)
{
int temp = data[pos1];
data[pos1] = data[pos2];
data[pos2] = temp;
}

Trying to increase the stack size due to an overflow, would be like buying more rubbish bins, when your bin is full instead of taking it to the dump.
Most probably you go into an endless recursion. Could you post your code?

You can use the following JVM options:
-Xms initial java heap size
-Xmx maximum java heap size
-Xss Set thread stack size
If you want to set these options by default in BlueJ, you need to do the following:
Find bluej.defs file
Find bluej.vm.args property (line) within that file
Add the option that you want in that line, i.e. bluej.vm.args = -Xmx512m to set the heap size to a maximum of 512 MB.
I hope this helps.

The stackoverflow error is usually because of a bad recursive call. Are you sure you're not doing anything wrong like specifying proper exit paths (aka terminating conditions ) for your recursion flow?

to me it looks like it's the partition that's bugged
private static int partition(int[] A, int first, int n )
{
int right = first + n-1;
int ls = first;
int pivot = A[right];//use right most for pivot
for(int i = first;i<right;i++)
{
if(A[i]<pivot){
swap(A,i,ls);
ls++;//inc after swap
}
}
swap(A,right,ls);
return ls;
}
I got this code from wikipedia

The simplest implementation of Quicksort is vulnerable to O(N) memory use in the worst case. It is possible to modify it to use O(log N) in the worst case, by only recursing into the smaller subarray and by turning the remaining recursion into a while loop:
//the following code probably contains of-by-one errors
quicksort(xs, begin, end):
while(not empty list){
mid = partition(xs, begin, end)
if( mid-begin < end-mid){
quicksort(xs, begin, mid)
end = mid
}else{
quicksort(xs, mid, end)
begin = mid
}