Quick Sort Beginner Java

/*
 * Kyle Arthur Benzle
 * Lab 3
 * Oct. 30 9
 * Tagore
 */ 
import java.io.*;
import java.util.Scanner;
//Lab3 Class
public class Lab3 {
	public static void main(String[] args) throws IOException {
		//getting the user input from scanner, calling it "input"
		System.out.print("Name of file with array: ");
		Scanner readIn = new Scanner(System.in);
		String input = readIn.nextLine();
		//testScan1(input);
		
	}
	private String filename;
	//Calling this method testScan1 to read in the file info, taking in the filename as a parameter
	//public static void testScan1(String filename) {
		//Calling the file, file
		File file = new File(filename);
		Scanner scan;
		//making an int array called array of course.
		int[] array;{
		//using try/catch block to read from file, (Incase file is not found)
		try {
			//Lab assignment never said anything about what our input would be, I assume there will be 6 intergers like in the example, I think this is a fair assumption as it was not defined that we had to have the array length variable.
			array = new int[5];
			//scan to get file info
			scan = new Scanner(file);
			//if, the file name was not found, give this error
		} catch (java.io.FileNotFoundException e) {
			System.out.println("couldn't open. file not found ");
			
		}
		int i =0;
		//Now, finally, reading the file, while loop for as long as i is < file.length. 
		while (scan.hasNext()) {
			//the below comment is for printing the array, Just a test.
			//for (i = 0; i <= file.length(); ++i) 
			//setting each of the array element = to the next number in the file
				array[i] = Integer.parseInt(scan.next());
				//i++;
				i++;}
		//Calling the quicksort method with array, 0 for left and length-1 for right
				quickSort(array, 0, array.length-1);
				//calling partition method, with the same as the above
			partition(array, 0, array.length-1);
			//Just another test, trying to print array
				//System.out.println(java.util.Arrays.toString(array));
			//printing array
				//for (int j = 0; j < array.length; ++j)
				//System.out.println(array[i] + " ");
			}
		
	
//partition method
	static int partition(int[] arr, int left, int right) {
		int i = left;
		int j = right;
		int tmp;
		//making pivot = to the last element of array
		int pivot = arr.length-1;
		//3 while loops to move the left and right toward each other, until the can be swapped below.
		while (i <= j) {
			while (arr[i] < pivot)
				i++;
			while (arr[j] > pivot)
				j--;
			if (i <= j) {
				tmp = arr[i];
				arr[i] = arr[j];
				arr[j] = tmp;
				i++;
				j--;
			}
		}
		return i;
	}
//QuickSort Method, taking in arr, left and right
	static void quickSort(int arr[], int left, int right) {
		//Setting index = to the partition with parameters of array and left and right, when that method runs
		if (left>=1) return;
		int index = partition(arr, left, right); //ERROR this line is giving an exception when the program runs, a stack overflow. 
		//if (left < (index - 1)) ; 
		
		//Call quickSort recursively with the new right
		quickSort(arr, left, index); //ERROR, this line is giving the overflow error I think you were talking about in class, that a lot of people were having trouble with. 
		//if (index < right) {
			//or call quickSort recursively with the old #'s
		quickSort(arr, index+1, 1);
	}}
	
			//Now, trying to print out the array. 
	//for (int j = 0; j < arr.length; ++j)
	//System.out.println(arr[j] + " ");
	//	}
	
                                              

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