算法01

来源：互联网发布：淘宝少发了一件怎么办编辑：程序博客网时间：2024/05/18 03:40

/** * 将数组s中的内容复制到数组d中,从from开始复制,from+i结束,发如果s数组长度不够,则补零 *  * @param d * @param s * @param from * @param i */public static void byteFirstCopy(byte d[], byte s[], int from, int i) {int j;for (j = 0; j < i; j++) {if (j < s.length) {d[j + from] = s[j];} else {d[j + from] = (byte) 0;}}}

/** * 将数组s中从from开始的maxlen个元素复制到数组d中,以0结束 *  * @param d * @param s * @param from * @param maxlen * @return */public static int arrayCopy(byte d[], byte s[], int from, int maxlen) {int i;for (i = 0; i < maxlen; i++)d[i + from] = s[i];return i;}

/** * 将整数转换为4个字节长度的数组 *  * @param longValue * @param arrayLen *            字节数组长度 * @return */public static byte[] long2Byte(long intValue) {byte[] b = new byte[4];int mark = 0xFF;//11111111for (int i = 0; i < 4; i++) {b[i] = (byte) (intValue >>> 8 * (3 - i) & mark);}return b;}

/** * 将整数转换成大小指定为arrayLen的字节数组，请注意value与arrayLen之间的关系，其是否存在越界。 *  * 请检查value的最大值为2的arrayLen*8次方减一， *  * 例如：arrayLen = 1；maxValue = 2^1*8-1=255; *  * @param value *            整数值 * @param arrayLen *            数组长度 * @return 该整形所对应的字节数组 */public static byte[] long2Byte(long value, int arrayLen) {byte[] bytes = new byte[arrayLen];if (arrayLen < 4) {for (int i = 0; i < arrayLen; i++) {bytes[i] = (byte) (value >>> 8 * (arrayLen - i - 1) & 0xFF);}return bytes;}byte[] temp = CmppTool.long2Byte(value);CmppTool.byteFirstCopy(bytes, temp, arrayLen - temp.length, temp.length);return bytes;}

/** * 将字节数组转化成整数 *  * @param b *            字节数组 * @return 字节数组所对应的整数 */public static long byte2Long(byte[] b) {long longValue = 0, mark = 0xFF;byte[] temp;int multiple = 0;if (b.length % 4 != 0) {multiple = b.length / 4 + 1;temp = new byte[4 * multiple];if (b.length > 4) {for (int i = 1; i < b.length; i++) {temp[4 * multiple - i] = b[b.length - i];}} else {for (int i = 0; i < b.length; i++) {longValue += (b[i] & mark) << (8 * (b.length - i - 1));}return longValue;}} else {temp = b;}for (int i = 0; i < temp.length; i++) {longValue += (temp[i] & mark) << (8 * (7 - i));}return longValue;}

0 0