问题描述

如果你有二进制数 10110，我怎样才能让它返回 5?例如一个数字，它告诉我们使用了多少位?下面列出了一些类似的例子:

If you have the binary number 10110 how can I get it to return 5? e.g a number that tells how many bits are used? There are some likewise examples listed below:

• 101 应该返回 3
• 000000011 应该返回 2
• 11100 应该返回 5
• 101010101 应该返回 9

如何在 Java 中以最简单的方式获得它?我想出了以下方法，但我可以做得更快吗:

How can this be obtained the easiest way in Java? I have come up with the following method but can i be done faster:

public static int getBitLength(int value)
{
    if (value == 0)
    {
        return 0;
    }
    int l = 1;
    if (value >>> 16 > 0) { value >>= 16; l += 16; }
    if (value >>> 8 > 0) { value >>= 8; l += 8; }
    if (value >>> 4 > 0) { value >>= 4; l += 4; }
    if (value >>> 2 > 0) { value >>= 2; l += 2; }
    if (value >>> 1 > 0) { value >>= 1; l += 1; }
    return l;
}

推荐答案

最简单?

32 - Integer.numberOfLeadingZeros(value)

如果您正在寻找算法，Java API 的实现者同意您的分而治之的位移方法:

If you are looking for algorithms, the implementors of the Java API agree with your divide-and-conquer bitshifting approach:

public static int numberOfLeadingZeros(int i) {
    if (i == 0)
        return 32;
    int n = 1;
    if (i >>> 16 == 0) { n += 16; i <<= 16; }
    if (i >>> 24 == 0) { n +=  8; i <<=  8; }
    if (i >>> 28 == 0) { n +=  4; i <<=  4; }
    if (i >>> 30 == 0) { n +=  2; i <<=  2; }
    n -= i >>> 31;
    return n;
}

编辑:提醒那些相信浮点计算准确性的人，运行以下测试工具:

Edit: As a reminder to those who trust in the accuracy of floating point calculations, run the following test harness:

public static void main(String[] args) {
    for (int i = 0; i < 64; i++) {
        long x = 1L << i;
        check(x);
        check(x-1);
    }
}

static void check(long x) {
    int correct = 64 - Long.numberOfLeadingZeros(x);
    int floated = (int) (1 + Math.floor(Math.log(x) / Math.log(2)));
    if (floated != correct) {
        System.out.println(Long.toString(x, 16) + " " + correct + " " + floated);
    }
}

第一个检测到的偏差是:

The first detected deviation is:

ffffffffffff 48 49

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