[FIXED] Self Define CompareTo(), why does the order of compareTo matters to the sorting order?

Issue

I have String array of [3,30,34,5,9] that I want to sort,

However, for value 3 and 30, 3 has to be bigger than 30

such that string a+b > b+c. if a = 3 and b = 30; ab>ba => 330>303 so that 3 should be bigger than 30;

So that my array [3,30,34,5,9] after sorting should be [9,5,34,3,30]

I found a class function that implements comparator as following:

private class LargerNumberComparator implements Comparator<String> {
    @Override
    public int compare(String a, String b) {
        String order1 = a + b;
        String order2 = b + a;
       return order1.compareTo(order2);
    }
}

In the above code, it will sort my string in ascending order such that original array: [3,30,34,5,9] will become [30,3,34,5,9]

However, if I change the return order1.compareTo(order2); to be return order2.compareTo(order1) that will change the sorting order of the array from ascending to descending;

Q:Why does the order to .compareTo changes the sorting order?

Solution

compare(a, b) returns a number that is negative if a < b, zero if a == b, and positive if a > b.

Let's assume compare(a, b) returns a negative number, so a < b.

Then order1.compareTo(order2) returned a negative number too, so order1 < order2.

But if you returned order2.compareTo(order1), well, we know order1 < order2. But that means order2 > order1, so order2.compareTo(order1) will return a positive number.

In general, it's really just that a < b is not the same thing as b < a -- it's the opposite.

Answered By - Louis Wasserman
Answer Checked By - Dawn Plyler (JavaFixing Volunteer)