转移Java BitSet

Question

我使用java.util.BitSet来存储密集的位向量。

我想实现一个将位向右移1的操作，类似于>>> on int。

是否有一个库函数可以移动BitSet ？

如果没有，有没有比下面更好的方法？

public static void logicalRightShift(BitSet bs) {
  for (int i = 0; (i = bs.nextSetBit(i)) >= 0;) {
    // i is the first bit in a run of set bits.

    // Set any bit to the left of the run.
    if (i != 0) { bs.set(i - 1); }

    // Now i is the index of the bit after the end of the run.
    i = bs.nextClearBit(i);  // nextClearBit never returns -1.
    // Clear the last bit of the run.
    bs.clear(i - 1);

    // 0000111100000...
    //     a   b
    // i starts off the loop at a, and ends the loop at b.
    // The mutations change the run to
    // 0001111000000...
  }
}

Answer 1

这应该够了吧：

BitSet shifted = bs.get(1, bs.length());

它会给你一个等于orginial的bitset，但没有最低位。

编辑：

将此概括为n位，

BitSet shifted = bs.get(n, Math.max(n, bs.length()));

Answer 2

请找到BitSet“左移”的代码块

/**
 * Shift the BitSet to left.<br>
 * For example : 0b10010 (=18) => 0b100100 (=36) (equivalent to multiplicate by 2)
 * @param bitSet
 * @return shifted bitSet
 */
public static BitSet leftShiftBitSet(BitSet bitSet) {
    final long maskOfCarry = 0x8000000000000000L;
    long[] aLong = bitSet.toLongArray();

    boolean carry = false;
    for (int i = 0; i < aLong.length; ++i) {
        if (carry) {
            carry = ((aLong[i] & maskOfCarry) != 0);
            aLong[i] <<= 1;
            ++aLong[i];
        } else {
            carry = ((aLong[i] & maskOfCarry) != 0);
            aLong[i] <<= 1;
        }
    }

    if (carry) {
        long[] tmp = new long[aLong.length + 1];
        System.arraycopy(aLong, 0, tmp, 0, aLong.length);
        ++tmp[aLong.length];
        aLong = tmp;
    }

    return BitSet.valueOf(aLong);
}

Answer 3

可能更有效的替代方案是使用底层的long []。

使用bitset.toLongArray()获取基础数据。 相应地移动那些长点，然后通过BitSet.valueOf(long[])创建一个新的BitSet你必须非常小心地移动底层长点，因为你必须取低位并将其转换为高位在数组的下一个长。

这应该允许您使用处理器本机的位移操作一次移动64位，而不是分别迭代每一位。

编辑：根据Louis Wasserman的评论。 这仅适用于Java 1.7 API。 当我写它时没有意识到。

Answer 4

这些函数分别模仿<<和>>>运算符。

/**
 * Shifts a BitSet n digits to the left. For example, 0b0110101 with n=2 becomes 0b10101.
 *
 * @param bits
 * @param n the shift distance.
 * @return
 */
public static BitSet shiftLeft(BitSet bits, int n) {
    if (n < 0)
        throw new IllegalArgumentException("'n' must be >= 0");
    if (n >= 64)
        throw new IllegalArgumentException("'n' must be < 64");

    long[] words = bits.toLongArray();

    // Do the shift
    for (int i = 0; i < words.length - 1; i++) {
        words[i] >>>= n; // Shift current word
        words[i] |= words[i + 1] << (64 - n); // Do the carry
    }
    words[words.length - 1] >>>= n; // shift [words.length-1] separately, since no carry

    return BitSet.valueOf(words);
}

/**
 * Shifts a BitSet n digits to the right. For example, 0b0110101 with n=2 becomes 0b000110101.
 *
 * @param bits
 * @param n the shift distance.
 * @return
 */
public static BitSet shiftRight(BitSet bits, int n) {
    if (n < 0)
        throw new IllegalArgumentException("'n' must be >= 0");
    if (n >= 64)
        throw new IllegalArgumentException("'n' must be < 64");

    long[] words = bits.toLongArray();

    // Expand array if there will be carry bits
    if (words[words.length - 1] >>> (64 - n) > 0) {
        long[] tmp = new long[words.length + 1];
        System.arraycopy(words, 0, tmp, 0, words.length);
        words = tmp;
    }

    // Do the shift
    for (int i = words.length - 1; i > 0; i--) {
        words[i] <<= n; // Shift current word
        words[i] |= words[i - 1] >>> (64 - n); // Do the carry
    }
    words[0] <<= n; // shift [0] separately, since no carry

    return BitSet.valueOf(words);
}

Answer 5

您可以使用BigInteger而不是BitSet 。 BigInteger已经有ShiftRight和ShiftLeft。

Answer 6

您可以查看BitSet toLongArray和valueOf(long[]) 。
基本上得到long数组，移位long s并从移位数组构造一个新的BitSet 。

Answer 7

为了获得更好的性能，您可以扩展java.util.BitSet实现并避免不必要的数组复制。 这是实现（我基本上重用了Jeff Piersol实现）：

package first.specific.structure;

import java.lang.reflect.Field;
import java.util.BitSet;

public class BitSetMut extends BitSet {

    private long[] words;
    private static Field wordsField;

    static {
        try {
            wordsField = BitSet.class.getDeclaredField("words");
            wordsField.setAccessible(true);
        } catch (NoSuchFieldException e) {
            throw new IllegalStateException(e);
        }
    }

    public BitSetMut(final int regLength) {
        super(regLength);
        try {
            words = (long[]) wordsField.get(this);
        } catch (IllegalAccessException e) {
            throw new IllegalStateException(e);
        }
    }

    public void shiftRight(int n) {
        if (n < 0)
            throw new IllegalArgumentException("'n' must be >= 0");
        if (n >= 64)
            throw new IllegalArgumentException("'n' must be < 64");

        if (words.length > 0) {
            ensureCapacity(n);

            // Do the shift
            for (int i = words.length - 1; i > 0; i--) {
                words[i] <<= n; // Shift current word
                words[i] |= words[i - 1] >>> (64 - n); // Do the carry
            }
            words[0] <<= n; // shift [0] separately, since no carry
            // recalculateWordInUse() is unnecessary
        }
    }

    private void ensureCapacity(final int n) {
        if (words[words.length - 1] >>> n > 0) {
            long[] tmp = new long[words.length + 3];
            System.arraycopy(words, 0, tmp, 0, words.length);
            words = tmp;
            try {
                wordsField.set(this, tmp);
            } catch (IllegalAccessException e) {
                throw new IllegalStateException(e);
            }
        }
    }
}

Answer 8

使用java SE8，它可以实现更简洁的方式：

BitSet b = new BitSet();
b.set(1, 3);
BitSet shifted = BitSet.valueOf(Arrays.stream(
       b.toLongArray()).map(v -> v << 1).toArray());

我试图弄清楚如何使用LongBuffer来做到这一点，但并没有让它发挥作用。 希望熟悉低级编程的人可以指出解决方案。

提前致谢！！！