如何将一串零和一转换为位集? [英] How to translate a string of zeros and ones into bitset?

问题描述

因此，我有以下代码片段将字符串转换为位集.

So I have this code snippet to translate a string into bitset.

    String huffmancode = "0010110100";
    char[] ch = huffmancode.toCharArray();

    BitSet bs = new BitSet();
    for (int i = 0; i < ch.length; i++)  {
        if (ch[i] == '1') {
            bs.set(i);
        }
    }

我的问题是，在霍夫曼代码的第一个和最后一个索引为0的情况下，如何确定位集的边界/大小/长度?

My question is how to determine the boundary / size / length of the bitset given that the first and the last indexes of huffman code were 0's ?

推荐答案

以下位集按顺序包含[0,1]，以下代码的最后一行将打印出2，即位集的长度.

The following bitset contains [0,1] in order and the last line of the following code prints out 2, the length of the bitset.

BitSet bs = new BitSet();

bs.set(0, false);
bs.set(1, true);

System.out.println(bs.length());

这篇关于如何将一串零和一转换为位集?的文章就介绍到这了，希望我们推荐的答案对大家有所帮助，也希望大家多多支持IT屋！