CODEWITHSUNDEEP
javaarrayssortingrecursion
user1908627
user1908627

Reputation: 21

Sorting an array recursively in Java with even numbers appearing in front of array.

I'm using a method to sort out an array with even numbers coming out in the front and odd numbers in the back of the array. My assignment dictates that i accomplish this using recursion. When I try to print the sorted out array it just prints out the array unsorted. What am i doing wrong?

The left variable starts at index 0 and the right variable starts at the end of the index. They are then both compared, if left is an odd number and right is an even number then they swap values. If left turns out to be an even number then no swap takes place and it points at the next index in the array. If right turns out to be an odd number then no swap takes place and it points at the next index in the array. It does this until all the even numbers are on the right of the array.

I'm getting a

"Exception in thread "main" java.lang.StackOverflowError".

import java.util.*;
public class Problem2{

    //i=left 
    //j=right
    //first i tried to shift the whole thing
    //have all even numbers pop to the front of array when even
    public static int[] callme(int[] arry, int left, int right){
        int temp;
        if(left>=right)       //base case, return array
            return arry; 
        else if(arry[left]%2!=0 && arry[right]%2==0){//if match, do the swap
            temp=arry[left];
            arry[left]=arry[right];
            arry[right]=temp;   
            return callme(arry, left++, right--);
        }
        else{
            if(arry[right]%2!=0){//if right side is on odd #, then decrease index
                return callme(arry, left, right--);
            }
            if(arry[left]%2==0){//if left side is on even #, then increase index
                return callme(arry, left++, right);
            }
        } 
        return arry;
    }

    public static void main(String[] args){

        //int index=0;
        int[] arry={3,5,6,8};

        int[] newarry=callme(arry, 0, arry.length-1);
        System.out.print("The new sorted array is: ");
        for(int i=0; i<newarry.length;i++){
            System.out.print(newarry[i]+" ");
        } 
    }
}

Upvotes: 1

Views: 3308

Answers (2)

ajb
ajb

Reputation: 31689

left++ is not the same as left + 1. If you want to call the method recursively with a left argument that is one higher, then call it with left + 1.

In this statement:

  return callme(arry, left++, right--);

the way it works is this: The program saves the current value of left, adds 1 to left, but then uses the value saved before it was incremented as the argument to the recursive call. The right-- argument works the same way. So when callme calls itself, it's calling itself with the exact same arguments it was called with. So you never get to the base case.

(Something else you should know about recursion: Each time a recursive method calls itself, it will have its own copy of the local variables, including the parameters. So even though the first method increases left, that has no impact on what left will be when it's called recursively, because the recursive methods have their own left.)

Upvotes: 1

user3163916
user3163916

Reputation: 328

If you just want the even numbers to be on the top, no need to do left and right.

You could do:

int temp;

for (int i = 0; i < array.length; i++)
  if (array[i] % 2)
    for (int j = i + 1; j < array.length; j++)
      if !(array[j] % 2) {
        temp = array[j];
        array[j] = array[i];
        array[i] = temp;
        j = array.length;
      }

Upvotes: 0

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