任意进制字符串的加减算法

Rick Murphy — Wed, 08 Apr 2009 14:28:00 GMT

private static final char[] nchars = new char[] {

    '0', '1', '2', '3', '4', '5', '6', '7', '8', '9',

    'A', 'B', 'C', 'D', 'E', 'F', 'G', 'H', 'I', 'J',

    'K', 'L', 'M', 'N', 'O', 'P', 'Q', 'R', 'S', 'T',

    'U', 'V', 'W', 'X', 'Y', 'Z' };

    public static String getNextString(String source, int step) {

        if (source == null || source.trim().length() == 0)

            return "0";

        char[] temp = source.trim().toCharArray();

        int capacity = nchars.length;

        int carry = step;

        for (int i = temp.length - 1; i >= 0; i--) {

            char c = temp[i];

            // get index of current char

            int index = 0;

            for (int n = 0; n < nchars.length; n++)

                if (c == nchars[n]) {

                    index = n;

                    break;

                }

            // calc carry and mod

            int total = index + carry;

            index = total % capacity;

            carry = total / capacity;

            // if the mod value is nagative, borrow 1 from left

            if (index < 0) {

                index = capacity + index;

                carry--;

            }

            temp[i] = nchars[index];

            // out of range

            if (carry != 0 && i == 0)

                throw new RuntimeException(String.format(

                "The number is out of range : %1$s + (%2$d)", source,

                step));

            if (carry == 0)

                break;

        }

        return new String(temp);

    }

Rick Murphy 2009-04-08 22:28 发表评论

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任意进制字符串的加减算法