英文:
Check if number is divisible by a power of two
问题
Find the highest power of two that divides x, a 64-bit integer, or return -1.
The zero case is not defined, since it divides any power of two, so your method can return any number.
I tried using the BigInteger.getLowestSetBit() for this, it returns the right answer but it's far from being optimal.
Example: Input -> output
- 3 -> -1
- 6 -> 1
- 4256 -> 5
英文:
Find the highest power of two that divides x, a 64-bit integer, or return -1.
The zero case is not defined, since it dives any power of two, so your method can return any number.
I tried using the BigInteger.getLowestSetBit() for this, it returns the right answer but it's far from being optimal.
Example: Input -> output
- 3 -> -1
- 6 -> 1
- 4256 -> 5
答案1
得分: 5
在 Long 类中，有一个便捷的静态函数 numberOfTrailingZeros，它几乎实现了你想要的功能，除了在输入不能被 2 整除时会返回 0（而不是 -1）。你可以用不同的方式处理这种情况。例如，扩展 @0x476f72616e 的答案：
if ((num & 0x1) == 0)
return Long.numberOfTrailingZeros(num);
else
return -1;
英文:
In the Long class, there is a handy static function numberOfTrailingZeros, which does almost what you want, except that it returns zero (instead of -1) when the input is not divisible by 2. You could handle that case in different ways. For example, extending the answer of @0x476f72616e
if ((num & 0x1) == 0)
return Long.numberOfTrailingZeros(num);
else
return -1;
答案2
得分: 2
一个可能的算法是：（伪代码）
使用一个计数器，将其设置为零，将数字放入一个整数变量 intvar 中
执行以下步骤：
- 右移（整数除以二）-> dividedvar
- 如果 dividedvar * 2 不等于 intvar，则设置 dividedvar = 0 /退出条件/
- 否则（intvar = dividedvar 且 计数器加一）
重复执行以上步骤，直到 dividedvar 不等于 0
尝试一下
英文:
one algorithm may be: (pseudocode)
use a counter set to zero, put number in a var intvar
do{
- shift right (integer divide by two) ->dividedvar
- if dividedvar*2 != intvar then dividedvar = 0 /exit condition/
- else (intvar = dividedvar and counter ++)
}
while dividedvar !=0
try it
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