Solution 1:

You can use Integer.parseInt with a radix of 2 (binary) to convert the binary string to an integer:

int charCode = Integer.parseInt(info, 2);

Then if you want the corresponding character as a string:

String str = new Character((char)charCode).toString();

Solution 2:

This is my one (Working fine on Java 8):

String input = "01110100"; // Binary input as String
StringBuilder sb = new StringBuilder(); // Some place to store the chars

Arrays.stream( // Create a Stream
    input.split("(?<=\\G.{8})") // Splits the input string into 8-char-sections (Since a char has 8 bits = 1 byte)
).forEach(s -> // Go through each 8-char-section...
    sb.append((char) Integer.parseInt(s, 2)) // ...and turn it into an int and then to a char
);

String output = sb.toString(); // Output text (t)

and the compressed method printing to console:

Arrays.stream(input.split("(?<=\\G.{8})")).forEach(s -> System.out.print((char) Integer.parseInt(s, 2))); 
System.out.print('\n');

I am sure there are "better" ways to do this but this is the smallest one you can probably get.

Solution 3:

I know the OP stated that their binary was in a String format but for the sake of completeness I thought I would add a solution to convert directly from a byte[] to an alphabetic String representation.

As casablanca stated you basically need to obtain the numerical representation of the alphabetic character. If you are trying to convert anything longer than a single character it will probably come as a byte[] and instead of converting that to a string and then using a for loop to append the characters of each byte you can use ByteBuffer and CharBuffer to do the lifting for you:

public static String bytesToAlphabeticString(byte[] bytes) {
    CharBuffer cb = ByteBuffer.wrap(bytes).asCharBuffer();
    return cb.toString();
}

N.B. Uses UTF char set

Alternatively using the String constructor:

String text = new String(bytes, 0, bytes.length, "ASCII");