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I've attempted to write an insertion sort and when looked up on the internet, I couldn't understand the code.

Here's what I did:

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

Inside my main method, I wrote this:

public class TestAlgos { public static void main (String args []) { int array[] = {2,5,3,6,8,0,4,2,4,6,1,4,6,9,3}; InsertionSort.sort(array); System.out.println(array); } }

But when run, I got this output (I used eclipse by the way):

[I@ed17bee

Thus I searched online for solutions and found this code on a website.

while(i > 0 && Array[i] > key) { Array[i + 1] = Array[i]; i = i - 1; } Array[i + 1] = key;

I don't know whether my code is wrong. Please explain.

Also, my second question is why instead of error, a number-string thingy is displayed as output.

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  • there are a lot of insertion sort variations. If your approach is different it doesn't mean it's worse. Are you getting the desired output with your algorithm? Commented Sep 18, 2018 at 15:51
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    Regarding your second question.. stackoverflow.com/questions/409784/… Commented Sep 18, 2018 at 16:14

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There is nothing wrong with your code since it's a valid Insertion sort implementation. Still the second code you've quoted is a better approach just because it makes less insertions.

Main difference is as follows: Your algorithm switches values until there are no more operations available/required for the current index's value.

Quoted algorithm shifts down the indexes of the values until no more operations are available/required and then places the value stored in key variable under the current index.

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