code: String[] halfMatch(String text1, String text2) {
          if (options.getTimeout() <= 0) {
              // Don't risk returning a non-optimal diff if we have unlimited time.
              return null;
          }
          String longtext = text1.length() > text2.length() ? text1 : text2;
          String shorttext = text1.length() > text2.length() ? text2 : text1;
          if (longtext.length() < 4 || shorttext.length() * 2 < longtext.length()) {
              return null;  // Pointless.
          }
  
          // First check if the second quarter is the seed for a half-match.
          String[] hm1 = halfMatchI(longtext, shorttext,
                                         (longtext.length() + 3) / 4);
          // Check again based on the third quarter.
          String[] hm2 = halfMatchI(longtext, shorttext,
                                         (longtext.length() + 1) / 2);
          String[] hm;
          if (hm1 == null && hm2 == null) {
              return null;
          } else if (hm2 == null) {
              hm = hm1;
          } else if (hm1 == null) {
              hm = hm2;
          } else {
              // Both matched.  Select the longest.
              hm = hm1[4].length() > hm2[4].length() ? hm1 : hm2;
          }
  
          // A half-match was found, sort out the return data.
          if (text1.length() > text2.length()) {
              return hm;
              //return new String[]{hm[0], hm[1], hm[2], hm[3], hm[4]};
          } else {
              return new String[]{hm[2], hm[3], hm[0], hm[1], hm[4]};
          }
      }
docs: Do the two texts share a substring which is at least half the length of
  the longer text?
  This speedup can produce non-minimal diffs.
  @param text1 First string.
  @param text2 Second string.
  @return Five element String array, containing the prefix of text1, the
  suffix of text1, the prefix of text2, the suffix of text2 and the
  common middle.  Or null if there was no match.