Java split() is returning an empty first element

Question

I have a string[] = [5 5, 1 2 N, LMLMLMLMM, 3 3 E, MMRMMRMRRM]

When I split the 2nd and 4th elements. I get

[, L, M, L, M, L, M, L, M, M]

[, M, M, R, M, M, R, M, R, R, M]

import java.io.*;

public class Instruction {
public String[] instructionList;
public String filePath;

public Instruction(String fileName) {
    this.filePath = fileName;
}

public String[] readFile() throws IOException {
    FileInputStream in = new FileInputStream(this.filePath);
    BufferedReader br = new BufferedReader(new InputStreamReader(in));

    int n = 5;
    instructionList = new String[n];

    for (int j = 0; j < instructionList.length; j++) {
        instructionList[j] = br.readLine();
    }
    in.close(); 
    return instructionList;
}}

import java.util.Arrays;
public class RoverCommand {

public static void main(String[] args) throws Exception {

//Create new Instruction object with directions.txt.                
    Instruction directions = new Instruction("directions.txt");
    String[] instructions = directions.readFile();              
    String roverInstructions = Arrays.toString(instructions[2].split(""));
    System.out.println(roverInstructions);

}

I've tried replacing the empty space, to no avail. How can I split() without returning this empty first element?

Answer 1

String.split() takes a regular expression so it may not be operating the way you are expecting it, however if you wanted to use it you could do this:

System.out.println(Arrays.toString("LMLMLMLMM".split("(?!^)")));

Which outputs this:

[L, M, L, M, L, M, L, M, M]

Here is a explanation of the regular expression:

(?!^) Negative Lookahead - Assert that it is impossible to match the regex below
   ^ assert position at start of the string

This will give you the same output though:

System.out.println(Arrays.toString("LMLMLMLMM".toCharArray()));

In this case I would advocate for toCharArray() both will work with double byte chars, so in the end it comes down to readability.