Question

如何有效地计算整数的二进制表示中的尾随零的数量？

Answer 1

这是一个很好，快速和简单的实现：

public static int NumberOfTrailingZeros(int i)
{
    return _lookup[(i & -i) % 37];
}

private static readonly int[] _lookup =
    {
        32, 0, 1, 26, 2, 23, 27, 0, 3, 16, 24, 30, 28, 11, 0, 13, 4, 7, 17,
        0, 25, 22, 31, 15, 29, 10, 12, 6, 0, 21, 14, 9, 5, 20, 8, 19, 18
    };

（取自http://graphics.stanford.edu/~seander/bithacks.html#ZerosOnRightModLookup。）

Answer 2

只需从第一个数字开始制作一个面具并继续移动直到它找到一些东西：

public static int numTrailingBinaryZeros(int n)
{
    int mask = 1;
    for (int i = 0; i < 32; i++, mask <<= 1)
        if ((n & mask) != 0)
            return i;

    return 32;
}

Answer 3

在java中它的实现方式如下：



  public static int numberOfTrailingZeros(int i) {
        // HD, Figure 5-14
        int y;
        if (i == 0) return 32;
        int n = 31;
        y = i <<16; if (y != 0) { n = n -16; i = y; }
        y = i << 8; if (y != 0) { n = n - 8; i = y; }
        y = i << 4; if (y != 0) { n = n - 4; i = y; }
        y = i << 2; if (y != 0) { n = n - 2; i = y; }
        return n - ((i << 1) >>> 31);
    }

public static int numberOfTrailingZeros(int i) { // HD, Figure 5-14 int y; if (i == 0) return 32; int n = 31; y = i <<16; if (y != 0) { n = n -16; i = y; } y = i << 8; if (y != 0) { n = n - 8; i = y; } y = i << 4; if (y != 0) { n = n - 4; i = y; } y = i << 2; if (y != 0) { n = n - 2; i = y; } return n - ((i << 1) >>> 31); } 我认为这个想法来自“Hacker's Delight”

Answer 4

int trailingZeros(int n) {
    if(!n)
        return 32;

    for(int s = 0; !(n & 1); s++)
        n >>= 1;
    return s;
}

Answer 5

Java版本的此变体使用了更多的参数来删除最后的条件测试：

    public static uint Ctz(uint num)
    {
        if (num == 0) return 32;    // optional, otherwise returns 0

        uint tmp;
        uint res = 0;
        num &= (uint)-(int)num;  // Isolate bit
        tmp = num >> 16; if (tmp != 0) { num = tmp; res += 16; }
        tmp = num >> 8;  if (tmp != 0) { num = tmp; res += 8; }
        tmp = num >> 4;  if (tmp != 0) { num = tmp; res += 4; }
        return res + ((num >> 1) - (num >> 3));
    }

如何使用C＃计算整数的二进制表示中的尾随零

5 个答案: