CODEWITHSUNDEEP
javaexceptionquicksort
DemCodeLines
DemCodeLines

Reputation: 1920

Quicksort Array Code

public class Main
{

  public static void main (String args[])
  {
    int nums[]= {2, 3, 4, 5, 6, 7, 8, 9, 0, 1};
    for (int index = 0; index < 10; index++)
      System.out.print("  " +nums[index]);;
    System.out.println();

    quickSort(nums,0,9);
    for (int index = 0; index < 10; index++)
      System.out.print("  " +nums[index]);
    System.out.println();
  }


  private static void swap(int a[], int lft, int rt)
  {
    int temp;
    temp = a[lft];
    a[lft] = a[rt];
    a[rt] = temp;
  }


  public static int pivot(int firstpl, int lastpl)
  {
    if(firstpl >= lastpl)
      return -1;
    else
      return firstpl;
  }



  private static void quickSort(int a[], int first, int last)
  {
    int left,right;  
    int pivindex = pivot(first,last);
    if(pivindex >= 0)
    {
      left = pivindex +1;
      right = last;
      do
      {
        while ( a[left] < a[pivindex] && left <= right )
          left++;
        while (a[right] > a[pivindex])
          right--;
        if (right > left)
          swap(a,left,right);
      }
      while(left < right);

       swap (a, pivindex, right);
       quickSort( a, first, right-1);
       quickSort(a, right+1, last);

    }
  }      
}

That is the code for a quick sort. Everything works right if the array's 0 position number is "2" but if it is anything else, it throws an exception.

For example, if I put this in for array:

5 6 8 9 7 8 0 1 8 2

It gives the following error:

Exception in thread "main" java.lang.ArrayIndexOutOfBoundsException: 10
    at Main.quickSort(Main.java:48)
    at Main.quickSort(Main.java:59)
    at Main.quickSort(Main.java:59)
    at Main.main(Main.java:12)

Please help!

Upvotes: 1

Views: 11717

Answers (2)

Russell Zahniser
Russell Zahniser

Reputation: 16364

Suppose that first = 8 and last = 9 in one of the calls to quickSort(). Suppose also that a[8] > a[9]. Think about what will happen in the left++ loop...

(Remember, the a[left] < a[pivindex] part of the condition is checked before the left <= right part)

OK, try walking through the method:

Starting with first = 8, last = 9, I end up with pivindex = 8, left = 9, right = 9. I check the while condition:

while(a[9] < a[8] && 9 <= 9)

It's true, so I increment left. Now left = 10 and right = 9. I check the while condition again...

while(a[10] < a[8] && 10 <= 9)

... and go off the end of the array. Now, if only I had checked that 10 <= 9 before looking at a[left]!

OK, if you're still stuck I'll just come right out and tell you. The problem in that loop is that you are trying to access a[left] before checking if left is too big. You can fix the problem in one of two ways:

  1. Switch the order of the conditions, so that you check left first, or
  2. Change the <= to a <, so that left never gets bigger than right.

Upvotes: 2

jefflunt
jefflunt

Reputation: 33954

The problem is a simple one: you're inadvertently trying to access a 10-element array with index of 10, when the only valid indexes are 0-9.

This is happening on line #48 (or possibly 47, or 49). You're probably accidentally adding +1 to an index when you don't realize it, and exceeding the bounds of the array.

Figure out why your special case is happening, and you'll be home.

Upvotes: 2

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