Method diff_halfMatch

src/burp/Utils/diff_match_patch.java:683–719

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, th

(String text1, String text2)

   * prefix of text2, the suffix of text2 and the common middle.  Or null if there was no match.
   */
  public String[] diff_halfMatch(String text1, String text2) {
    if (Diff_Timeout <= 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 = diff_halfMatchI(longtext, shorttext,
        (longtext.length() + 3) / 4);
    // Check again based on the third quarter.
    String[] hm2 = diff_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]};
    }
  }

  /**
   * Does a substring of shorttext exist within longtext such that the