代表数字7的8位像这样:
00000111
设置了三个比特。
确定32位整数中设置位数的算法是什么?
代表数字7的8位像这样:
00000111
设置了三个比特。
确定32位整数中设置位数的算法是什么?
当前回答
一个快速的c#解决方案,使用预先计算的字节位计数表,并根据输入大小进行分支。
public static class BitCount
{
public static uint GetSetBitsCount(uint n)
{
var counts = BYTE_BIT_COUNTS;
return n <= 0xff ? counts[n]
: n <= 0xffff ? counts[n & 0xff] + counts[n >> 8]
: n <= 0xffffff ? counts[n & 0xff] + counts[(n >> 8) & 0xff] + counts[(n >> 16) & 0xff]
: counts[n & 0xff] + counts[(n >> 8) & 0xff] + counts[(n >> 16) & 0xff] + counts[(n >> 24) & 0xff];
}
public static readonly uint[] BYTE_BIT_COUNTS =
{
0, 1, 1, 2, 1, 2, 2, 3, 1, 2, 2, 3, 2, 3, 3, 4,
1, 2, 2, 3, 2, 3, 3, 4, 2, 3, 3, 4, 3, 4, 4, 5,
1, 2, 2, 3, 2, 3, 3, 4, 2, 3, 3, 4, 3, 4, 4, 5,
2, 3, 3, 4, 3, 4, 4, 5, 3, 4, 4, 5, 4, 5, 5, 6,
1, 2, 2, 3, 2, 3, 3, 4, 2, 3, 3, 4, 3, 4, 4, 5,
2, 3, 3, 4, 3, 4, 4, 5, 3, 4, 4, 5, 4, 5, 5, 6,
2, 3, 3, 4, 3, 4, 4, 5, 3, 4, 4, 5, 4, 5, 5, 6,
3, 4, 4, 5, 4, 5, 5, 6, 4, 5, 5, 6, 5, 6, 6, 7,
1, 2, 2, 3, 2, 3, 3, 4, 2, 3, 3, 4, 3, 4, 4, 5,
2, 3, 3, 4, 3, 4, 4, 5, 3, 4, 4, 5, 4, 5, 5, 6,
2, 3, 3, 4, 3, 4, 4, 5, 3, 4, 4, 5, 4, 5, 5, 6,
3, 4, 4, 5, 4, 5, 5, 6, 4, 5, 5, 6, 5, 6, 6, 7,
2, 3, 3, 4, 3, 4, 4, 5, 3, 4, 4, 5, 4, 5, 5, 6,
3, 4, 4, 5, 4, 5, 5, 6, 4, 5, 5, 6, 5, 6, 6, 7,
3, 4, 4, 5, 4, 5, 5, 6, 4, 5, 5, 6, 5, 6, 6, 7,
4, 5, 5, 6, 5, 6, 6, 7, 5, 6, 6, 7, 6, 7, 7, 8
};
}
其他回答
32位还是32位?我只是在阅读了“破解编码面试”第4版练习5.5(第5章:位操作)后,在Java中使用了这种方法。如果最小有效位是1个增量计数,则右移该整数。
public static int bitCount( int n){
int count = 0;
for (int i=n; i!=0; i = i >> 1){
count += i & 1;
}
return count;
}
我认为这个比常数0x33333333的解更直观,不管它们有多快。这取决于你对“最佳算法”的定义。
我发现了一个在数组中使用SIMD指令(SSSE3和AVX2)的位计数实现。它的性能比使用__popcnt64内禀函数要好2-2.5倍。
SSSE3版:
#include <smmintrin.h>
#include <stdint.h>
const __m128i Z = _mm_set1_epi8(0x0);
const __m128i F = _mm_set1_epi8(0xF);
//Vector with pre-calculated bit count:
const __m128i T = _mm_setr_epi8(0, 1, 1, 2, 1, 2, 2, 3, 1, 2, 2, 3, 2, 3, 3, 4);
uint64_t BitCount(const uint8_t * src, size_t size)
{
__m128i _sum = _mm128_setzero_si128();
for (size_t i = 0; i < size; i += 16)
{
//load 16-byte vector
__m128i _src = _mm_loadu_si128((__m128i*)(src + i));
//get low 4 bit for every byte in vector
__m128i lo = _mm_and_si128(_src, F);
//sum precalculated value from T
_sum = _mm_add_epi64(_sum, _mm_sad_epu8(Z, _mm_shuffle_epi8(T, lo)));
//get high 4 bit for every byte in vector
__m128i hi = _mm_and_si128(_mm_srli_epi16(_src, 4), F);
//sum precalculated value from T
_sum = _mm_add_epi64(_sum, _mm_sad_epu8(Z, _mm_shuffle_epi8(T, hi)));
}
uint64_t sum[2];
_mm_storeu_si128((__m128i*)sum, _sum);
return sum[0] + sum[1];
}
AVX2 版本:
#include <immintrin.h>
#include <stdint.h>
const __m256i Z = _mm256_set1_epi8(0x0);
const __m256i F = _mm256_set1_epi8(0xF);
//Vector with pre-calculated bit count:
const __m256i T = _mm256_setr_epi8(0, 1, 1, 2, 1, 2, 2, 3, 1, 2, 2, 3, 2, 3, 3, 4,
0, 1, 1, 2, 1, 2, 2, 3, 1, 2, 2, 3, 2, 3, 3, 4);
uint64_t BitCount(const uint8_t * src, size_t size)
{
__m256i _sum = _mm256_setzero_si256();
for (size_t i = 0; i < size; i += 32)
{
//load 32-byte vector
__m256i _src = _mm256_loadu_si256((__m256i*)(src + i));
//get low 4 bit for every byte in vector
__m256i lo = _mm256_and_si256(_src, F);
//sum precalculated value from T
_sum = _mm256_add_epi64(_sum, _mm256_sad_epu8(Z, _mm256_shuffle_epi8(T, lo)));
//get high 4 bit for every byte in vector
__m256i hi = _mm256_and_si256(_mm256_srli_epi16(_src, 4), F);
//sum precalculated value from T
_sum = _mm256_add_epi64(_sum, _mm256_sad_epu8(Z, _mm256_shuffle_epi8(T, hi)));
}
uint64_t sum[4];
_mm256_storeu_si256((__m256i*)sum, _sum);
return sum[0] + sum[1] + sum[2] + sum[3];
}
Kotlin 1.4 之前
fun NumberOfSetBits(i: Int): Int {
var i = i
i -= (i ushr 1 and 0x55555555)
i = (i and 0x33333333) + (i ushr 2 and 0x33333333)
return (i + (i ushr 4) and 0x0F0F0F0F) * 0x01010101 ushr 24
}
这或多或少是上面那个答案的翻版。
它带有Java补丁,然后使用IntelliJ IDEA Community Edition中的转换器进行转换
1.4及以上(截至2021-05-05 -未来可能会改变)。
fun NumberOfSetBits(i: Int): Int {
return i.countOneBits()
}
在底层,它使用Integer。bitCount如下所示:
@SinceKotlin("1.4")
@WasExperimental(ExperimentalStdlibApi::class)
@kotlin.internal.InlineOnly
public actual inline fun Int.countOneBits(): Int = Integer.bitCount(this)
def hammingWeight(n):
count = 0
while n:
if n&1:
count += 1
n >>= 1
return count
这是在golang中的实现
func CountBitSet(n int) int {
count := 0
for n > 0 {
count += n & 1
n >>= 1
}
return count
}