Sorting an int array by sign(positive, zero, negative) without sort method

Question

public static int[] sortBySign(int[] nums){
      int startIndex = 0;
      int endIndex = nums.length - 1;
 while(startIndex < endIndex){
    while(nums[startIndex] < 0){
          startIndex++;
    }
    while(nums[endIndex] > 0){
          endIndex--;
    }
          int temp = nums[startIndex];
          nums[startIndex] = nums[endIndex];
          nums[endIndex] = temp;
          startIndex++;
          endIndex--;
      }
    return nums;
  }

My code works for sorting positive and negative numbers but I'm not sure how to sort the zeroes as well. The negatives should be on the left side, zeroes in the middle, and positives on the right side. The order does not matter.

Have you stepped through the code in your debugger? This will help you determine on your own what needs to happen. — Jim Garrison
– Jim Garrison, Commented Jan 8, 2017 at 22:43

Sash Sinha · Accepted Answer · 2017-01-08 22:48:46Z

1

Using a auxiliary swap method you could handle the zeroes like so:

public static int[] sortBySign(int[] array) {
  int counter = 0;
  for (int i = 0; i < array.length; i++) {
    if (array[i] < 0) {
      swap(array, counter++, i);
    }
  }
  for (int i = counter; i < array.length; i++) {
    if (array[i] == 0) {
      swap(array, counter++, i);
    }
  }
  return array;
}

private static void swap(int array[], int index1, int index2) {
  int temp = array[index2];
  for (int i = index2; i > index1; i--) {
    array[i] = array[i - 1];
  }
  array[index1] = temp;
}

Try it here!

answered Jan 8, 2017 at 22:48

Sash Sinha

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Comments

Darshan Mehta · Accepted Answer · 2017-01-08 23:24:04Z

1

Actually, your code does not sort the positive numbers correctly, maybe because it's not doing enough number of iterations. To sort all the numbers (including zero), I would recommend falling back to bubble sort, e.g.:

public static void sort(int[] array) {
    for (int i = 0; i < array.length; i++) {
        for (int j = 1; j < (array.length - i); j++) {
            if (array[j - 1] > array[j]) {
                int temp = array[j - 1];
                array[j - 1] = array[j];
                array[j] = temp;
            }
        }
    }
}

Also, we don't need to return anything as the changes are being made to the actual array only.

Edit

Another solution to sort the array with one for loop, (i.e. O(n) complexity):

public static void sort(int[] array) {
    boolean continue = false;
    for (int i = 0; i < array.length - 1; i++) {
        if (array[i] < array[i + 1]) {
            int temp = array[i];
            array[i] = array[i + 1];
            array[i + 1] = temp; // swap values
            continue = true;
        }
        if (i == array.length - 2 && again) {
            i = 0;
            continue = false;
        }
    }
}

edited Jan 8, 2017 at 23:24

answered Jan 8, 2017 at 22:58

Darshan Mehta

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2 Comments

laura815 Over a year ago

Thanks. I was trying to implement it so that it's O(n).

Darshan Mehta Over a year ago

@laura815 I have added another solution.

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