String reversal function reverse

    /**
     * Causes this character sequence to be replaced by the reverse of
     * the sequence. If there are any surrogate pairs included in the
     * sequence, these are treated as single characters for the
     * reverse operation. Thus, the order of the high-low surrogates
     * is never reversed.
     *
     * Let <i>n</i> be the character length of this character sequence
     * (not the length in {@code char} values) just prior to
     * execution of the {@code reverse} method. Then the
     * character at index <i>k</i> in the new character sequence is
     * equal to the character at index <i>n-k-1</i> in the old
     * character sequence.
     *
     * <p>Note that the reverse operation may result in producing
     * surrogate pairs that were unpaired low-surrogates and
     * high-surrogates before the operation. For example, reversing
     * "\u005CuDC00\u005CuD800" produces "\u005CuD800\u005CuDC00" which is
     * a valid surrogate pair.
     *
     * @return  a reference to this object.
     */
    public AbstractStringBuilder reverse() {
        boolean hasSurrogates = false;
        int n = count - 1;
        for (int j = (n-1) >> 1; j >= 0; j--) {
            int k = n - j;
            char cj = value[j];
            char ck = value[k];
            value[j] = ck;
            value[k] = cj;
            if (Character.isSurrogate(cj) ||
                Character.isSurrogate(ck)) {
                hasSurrogates = true;
            }
        }
        if (hasSurrogates) {
            reverseAllValidSurrogatePairs();
        }
        return this;
    }

Source analysis:

  for (int j = (n-1) >> 1; j >= 0; j--) { }

>>On behalf of the right one, because during the reverse cycle time, only half of the elements reversed, for example: If you are 10, you will need to reverse five times, if the same is 9, you need to reverse five times.