Counting the Ones and Zeros of an String in Java, and putting them in order?-CodePudding

I do have a big problem coding something that should be easy:

I need to create a static public method which returns a string and also gets a binary number in form of an string.

public static String encode(String a) {

This method has to count the consecutive zeros and ones. If the number changes there has to be a blank. Also if the first number isn't Zero, the program has to just write a zero as first number.

Example: 11100110111 => "0 3 2 2 1 3"

Other Examples are:
encode("0") --> "1" 
encode("000") --> "3" 
encode("11") --> "0 2" 
encode("011") --> "1 2" 
encode("101") --> "0 1 1 1"

I have no idea how to do this, it would be so nice of someone could help! I will post my worthless piece of code here, its useless but eventually it contains some helpful information!

public class Functionality {
    public static void main(String[] args) {
        System.out.println(encode("111"));

    }

    public static String encode(String a) {
        int z = 0;
        int v = 0;
        String erg = "";

        while (a.length() == 0) {

            while (v == ErsterWert(a)) {
                z = z   1;
                subt(a);

            }

            String zx = String.valueOf(z);
            erg = erg   zx   " ";
            if (v == 0) {
                v = v   1;
            }

            if (v == 1) {
                v = 0;
            }

            z = 0;

        }
        String bg = "sds";
        return erg;
    }

    public static int ErsterWert(String a) {
        int f = a.charAt(0);
        return f;
    }

    public static String subt(String a) {
        a = a.substring(1);
        return a;
    }

}

CodePudding user response：

Here's a one-liner to impress your professor with:

public static String encode(String a) {
    return (a.startsWith("1") ? "0 " : "")  
        Arrays.stream(a.split("(?<=(.))(?!\\1)"))
          .map(String::length)
          .map(String::valueOf)
          .collect(Collectors.joining(" "));
}

This splits the string between characters that are different by using a lookbehind to capture the previous character and a negative lookahead for the back reference to that captured character.

These substrings are converted to their length, then the length as a String, finally being joined to a single string separated by a space.

The special case of the first character being a 1 is handled via a ternary.

CodePudding user response：

A recursive solution using String functions indexOf and substring may look like this:

we try to find the position of a digit X opposite to that in the beginning of the string to calculate the length of prefix.

public static String encode(String str) {
    if (str.startsWith("1")) {
        // append 0 at start, look for 0 after 1's
        return "0 "   encode(str, 0);
    }
    // string starts with 0, look for the first 1 after 0's
    return encode(str, 1);
}

// d - the digit to look for after current prefix 
public static String encode(String z, int d) {
    int pos = z.indexOf(String.valueOf(d));
    if (pos == -1) { // nothing found after current prefix
        return String.valueOf(z.length());
    }
    // append prefix length and toggle the number to look for
    return String.valueOf(pos)   " "   encode(z.substring(pos), 1 - d);
}

Tests:

for (String s : Arrays.asList("0", "1", "000", "11", "011", "101", "0011100001111100")) {
    System.out.println(s   " -> "   encode(s));
}

Output:

0 -> 1
1 -> 0 1
000 -> 3
11 -> 0 2
011 -> 1 2
101 -> 0 1 1 1
0011100001111100 -> 2 3 4 5 2