[Algorithm] N numbers with an interval of X

Problem description

The method solution takes a parameter of an integer x and a natural number n and returns a list with n numbers starting from x and incrementing by x. Take a look at the following conditions and create a method solution that satisfies the conditions.

conditions

--x is an integer greater than or equal to -10000000 and less than or equal to 10000000. --n is a natural number less than or equal to 1000.

Input / output example

Commentary

• The explanation is the code I wrote, so if you have a better algorithm, please share it!

class Solution {
    public long[] solution(int x, int n) {
        long[] result = new long[n];
        result[0] = x; //Start with x, so initialize x to Index 0
        
        //Since 0 has been initialized above, i starts from 1 and repeats until n.
        for (int i = 1; i < n; i++) {
            //Since it increases by x, the result Index: i-Value of 1+Do x.
            result[i] = result[i - 1] + x;
        }
        
        return result;
    }
}