f(a, b) returns string with only the characters found in both strings in the order of a

/* Write a function f(a, b) which takes two character string arguments and 
 * returns a string containing only the characters found in both strings in the
 * order of a. Write a version which is order N-squared and one which is order N.
 */
public class StrOverlap {

  public char[] overlapn2(char[] a, char[] b) {
    char [] o = new char[a.length];
    int k=0;
    for(int i=0; i<a.length; i++) {
      for(int j=0; j<b.length; j++) {
        if (a[i] == b[j]) {
          o[k++] = a[i];
          break; // go to the next 'a'
        }
      }
    }
    for(; k<o.length; k++) {
      o[k]= '\n';
    }
    return o;
  }

  public String overlapn(String b, String a, boolean showRepeats) {
    StringBuffer result = new StringBuffer();
    boolean[] asciiArr = new boolean[256]; // boolean ascii array (initially all false)
    char[] aarr = a.toCharArray();
    for (int i = 0; i < aarr.length; i++)
      asciiArr[aarr[i]] = true;
    char[] bbrr = b.toCharArray();
    for (int i = 0; i < bbrr.length; i++)
      if (asciiArr[bbrr[i]]) {
        result.append(bbrr[i]);
        if (!showRepeats) asciiArr[bbrr[i]] = false; // already printed
      }
    return result.toString();
  }

  public void init(String astr, String bstr) {
    char []a = new char[astr.length()];
    char []b = new char[bstr.length()];
    astr.getChars(0, astr.length(), a, 0);
    bstr.getChars(0, bstr.length(), b, 0);
    char [] arr = overlapn2(a, b);
    String stro = new String(arr);
    System.out.println("o: \n"+stro);
    String stron = overlapn(astr, bstr, true);
    System.out.println("o: \n"+stron);
  }

  static public void main(String[] args) {
    StrOverlap so = new StrOverlap();
    if(args.length>1) {
      so.init(args[0], args[1]);
    } else {
      System.out.println("usage: StrOverlap a b");
    }
  }

}