如何在保留长度的同时翻转 Java BitSet 中的所有位
[英]How to flip all the bits in Java BitSet while preserving its length
问题描述
How can I flip all the bits in a Java BitSet , while preserving its length?
如何翻转 Java BitSet中的所有位，同时保留其长度？
For example, given the following BitSet:例如，给定以下 BitSet：
BitSet.valueOf(new byte[] { 0b100111 })
// {0, 1, 2, 5}
Is there a simple method to flip all the bits, while preserving the BitSet length (6, in the above example)?是否有一种简单的方法来翻转所有位，同时保留 BitSet 长度（在上面的示例中为 6）？
I would like to get:我想得到：
BitSet.valueOf(new byte[] { 0b011000 })
// { 3, 4 }
1 个解决方案
解决方案1
0 2021-12-08 10:37:20
BitSet has a flip(from, to) method , which allows you to flip bits in a range: BitSet有一个flip(from, to)方法，它允许您翻转某个范围内的位：
yourBitSet.flip(0, length);
The question you need to answer, however, is what you actually mean by "length".但是，您需要回答的问题是“长度”的实际含义。
A
BitSethas thesize()method, but that reports the number of bits it has allocated space for - perhaps more than you think the size is.BitSet具有size()方法，但它报告它已分配空间的位数 - 可能比您认为的大小更多。It also has a
length()method, which reports the highest-set bit in the bit set - perhaps less than you think the length of the bitset is.它还有一个length()方法，它报告位集中设置的最高位 - 可能小于您认为的位集长度。
Assuming you're OK with using length() , you can use:假设您可以使用length() ，您可以使用：
yourBitSet.flip(0, yourBitSet.length());
Of course, since this clears the highest-set bit, yourBitSet will have a smaller length() afterwards.当然，由于这会清除设置的最高位， yourBitSet之后的length()会更小。
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