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
public class BinaryCounter {
private int N;
public BinaryCounter(int N) {
this.N = N;
}
public static void main(String[] args) {
BinaryCounter counter=new BinaryCounter(7);
System.out.println("Number of ones is "+ counter.count());
}
public int count(){
if(N<=0) return 0;
int counter=0;
int K = 0;
do{
K = biggestPowerOfTwoSmallerThan(N);
N = N-K;
counter++;
}while (N != 0);
return counter;
}
private int biggestPowerOfTwoSmallerThan(int N) {
if(N==1) return 1;
for(int i=0;i<N;i++){
if(Math.pow(2, i) > N){
int power = i-1;
return (int) Math.pow(2, power);
}
}
return 0;
}
}
其他回答
我提供了另一个未提及的算法，称为并行，从这里取。它的优点是它是通用的，这意味着代码对于8、16、32、64和128位大小是相同的。
我检查了它的值的正确性和时间的数量为2^26位大小为8,16,32和64位。请看下面的时间安排。
该算法是第一个代码片段。这里提到另外两个只是为了参考，因为我测试和比较了它们。
算法是用c++编写的，是通用的，但它可以很容易地应用到旧的C中。
#include <type_traits>
#include <cstdint>
template <typename IntT>
inline size_t PopCntParallel(IntT n) {
// https://graphics.stanford.edu/~seander/bithacks.html#CountBitsSetParallel
using T = std::make_unsigned_t<IntT>;
T v = T(n);
v = v - ((v >> 1) & (T)~(T)0/3); // temp
v = (v & (T)~(T)0/15*3) + ((v >> 2) & (T)~(T)0/15*3); // temp
v = (v + (v >> 4)) & (T)~(T)0/255*15; // temp
return size_t((T)(v * ((T)~(T)0/255)) >> (sizeof(T) - 1) * 8); // count
}
下面是我比较的两种算法。一个是带有循环的Kernighan简单方法，从这里开始。
template <typename IntT>
inline size_t PopCntKernighan(IntT n) {
// http://graphics.stanford.edu/~seander/bithacks.html#CountBitsSetKernighan
using T = std::make_unsigned_t<IntT>;
T v = T(n);
size_t c;
for (c = 0; v; ++c)
v &= v - 1; // Clear the least significant bit set
return c;
}
另一个是使用内置的__popcnt16()/__popcnt()/__popcnt64() MSVC的内在(doc在这里)。或CLang/GCC (doc)的__builtin_popcount。这个内在应该提供一个非常优化的版本，可能是硬件:
#ifdef _MSC_VER
// https://learn.microsoft.com/en-us/cpp/intrinsics/popcnt16-popcnt-popcnt64?view=msvc-160
#include <intrin.h>
#define popcnt16 __popcnt16
#define popcnt32 __popcnt
#define popcnt64 __popcnt64
#else
// https://gcc.gnu.org/onlinedocs/gcc/Other-Builtins.html
#define popcnt16 __builtin_popcount
#define popcnt32 __builtin_popcount
#define popcnt64 __builtin_popcountll
#endif
template <typename IntT>
inline size_t PopCntBuiltin(IntT n) {
using T = std::make_unsigned_t<IntT>;
T v = T(n);
if constexpr(sizeof(IntT) <= 2)
return popcnt16(uint16_t(v));
else if constexpr(sizeof(IntT) <= 4)
return popcnt32(uint32_t(v));
else if constexpr(sizeof(IntT) <= 8)
return popcnt64(uint64_t(v));
else
static_assert([]{ return false; }());
}
以下是计时，以纳秒为单位。所有的计时都是对2^26个随机数进行的。比较了所有三种算法的计时以及8、16、32和64之间的所有比特大小。总的来说，所有测试在我的机器上花费了16秒。使用了高分辨率时钟。
08 bit Builtin 8.2 ns
08 bit Parallel 8.2 ns
08 bit Kernighan 26.7 ns
16 bit Builtin 7.7 ns
16 bit Parallel 7.7 ns
16 bit Kernighan 39.7 ns
32 bit Builtin 7.0 ns
32 bit Parallel 7.0 ns
32 bit Kernighan 47.9 ns
64 bit Builtin 7.5 ns
64 bit Parallel 7.5 ns
64 bit Kernighan 59.4 ns
128 bit Builtin 7.8 ns
128 bit Parallel 13.8 ns
128 bit Kernighan 127.6 ns
可以看到，所提供的并行算法(在三个算法中排名第一)与MSVC /CLang的固有算法一样好。
作为参考，下面是我用来测试所有函数的速度/时间/正确性的完整代码。
作为奖励，这段代码(不像上面的短代码片段)也测试128位大小，但只在CLang/GCC下(不是MSVC)，因为它们有unsigned __int128。
在网上试试!
#include <type_traits>
#include <cstdint>
using std::size_t;
#if defined(_MSC_VER) && !defined(__clang__)
#define IS_MSVC 1
#else
#define IS_MSVC 0
#endif
#if IS_MSVC
#define HAS128 false
#else
using int128_t = __int128;
using uint128_t = unsigned __int128;
#define HAS128 true
#endif
template <typename T> struct UnSignedT { using type = std::make_unsigned_t<T>; };
#if HAS128
template <> struct UnSignedT<int128_t> { using type = uint128_t; };
template <> struct UnSignedT<uint128_t> { using type = uint128_t; };
#endif
template <typename T> using UnSigned = typename UnSignedT<T>::type;
template <typename IntT>
inline size_t PopCntParallel(IntT n) {
// https://graphics.stanford.edu/~seander/bithacks.html#CountBitsSetParallel
using T = UnSigned<IntT>;
T v = T(n);
v = v - ((v >> 1) & (T)~(T)0/3); // temp
v = (v & (T)~(T)0/15*3) + ((v >> 2) & (T)~(T)0/15*3); // temp
v = (v + (v >> 4)) & (T)~(T)0/255*15; // temp
return size_t((T)(v * ((T)~(T)0/255)) >> (sizeof(T) - 1) * 8); // count
}
template <typename IntT>
inline size_t PopCntKernighan(IntT n) {
// http://graphics.stanford.edu/~seander/bithacks.html#CountBitsSetKernighan
using T = UnSigned<IntT>;
T v = T(n);
size_t c;
for (c = 0; v; ++c)
v &= v - 1; // Clear the least significant bit set
return c;
}
#if IS_MSVC
// https://learn.microsoft.com/en-us/cpp/intrinsics/popcnt16-popcnt-popcnt64?view=msvc-160
#include <intrin.h>
#define popcnt16 __popcnt16
#define popcnt32 __popcnt
#define popcnt64 __popcnt64
#else
// https://gcc.gnu.org/onlinedocs/gcc/Other-Builtins.html
#define popcnt16 __builtin_popcount
#define popcnt32 __builtin_popcount
#define popcnt64 __builtin_popcountll
#endif
#define popcnt128(x) (popcnt64(uint64_t(x)) + popcnt64(uint64_t(x >> 64)))
template <typename IntT>
inline size_t PopCntBuiltin(IntT n) {
using T = UnSigned<IntT>;
T v = T(n);
if constexpr(sizeof(IntT) <= 2)
return popcnt16(uint16_t(v));
else if constexpr(sizeof(IntT) <= 4)
return popcnt32(uint32_t(v));
else if constexpr(sizeof(IntT) <= 8)
return popcnt64(uint64_t(v));
else if constexpr(sizeof(IntT) <= 16)
return popcnt128(uint128_t(v));
else
static_assert([]{ return false; }());
}
#include <random>
#include <vector>
#include <chrono>
#include <string>
#include <iostream>
#include <iomanip>
#include <map>
inline double Time() {
static auto const gtb = std::chrono::high_resolution_clock::now();
return std::chrono::duration_cast<std::chrono::duration<double>>(
std::chrono::high_resolution_clock::now() - gtb).count();
}
template <typename T, typename F>
void Test(std::string const & name, F f) {
std::mt19937_64 rng{123};
size_t constexpr bit_size = sizeof(T) * 8, ntests = 1 << 6, nnums = 1 << 14;
std::vector<T> nums(nnums);
for (size_t i = 0; i < nnums; ++i)
nums[i] = T(rng() % ~T(0));
static std::map<size_t, size_t> times;
double min_time = 1000;
for (size_t i = 0; i < ntests; ++i) {
double timer = Time();
size_t sum = 0;
for (size_t j = 0; j < nnums; j += 4)
sum += f(nums[j + 0]) + f(nums[j + 1]) + f(nums[j + 2]) + f(nums[j + 3]);
auto volatile vsum = sum;
min_time = std::min(min_time, (Time() - timer) / nnums);
if (times.count(bit_size) && times.at(bit_size) != sum)
std::cout << "Wrong bit cnt checksum!" << std::endl;
times[bit_size] = sum;
}
std::cout << std::setw(2) << std::setfill('0') << bit_size
<< " bit " << name << " " << std::fixed << std::setprecision(1)
<< min_time * 1000000000 << " ns" << std::endl;
}
int main() {
#define TEST(T) \
Test<T>("Builtin", PopCntBuiltin<T>); \
Test<T>("Parallel", PopCntParallel<T>); \
Test<T>("Kernighan", PopCntKernighan<T>); \
std::cout << std::endl;
TEST(uint8_t); TEST(uint16_t); TEST(uint32_t); TEST(uint64_t);
#if HAS128
TEST(uint128_t);
#endif
#undef TEST
}
对于JavaScript，你可以使用一个查找表来计算一个32位值的设置位的数量(这段代码可以很容易地翻译成C语言)。此外，添加了8位和16位版本，以供通过网络搜索查找的人使用。
const COUNT_BITS_TABLE = makeLookupTable() function makeLookupTable() { const table = new Uint8Array(256) for (let i = 0; i < 256; i++) { table[i] = (i & 1) + table[(i / 2) | 0]; } return table } function countOneBits32(n) { return COUNT_BITS_TABLE[n & 0xff] + COUNT_BITS_TABLE[(n >> 8) & 0xff] + COUNT_BITS_TABLE[(n >> 16) & 0xff] + COUNT_BITS_TABLE[(n >> 24) & 0xff]; } function countOneBits16(n) { return COUNT_BITS_TABLE[n & 0xff] + COUNT_BITS_TABLE[(n >> 8) & 0xff] } function countOneBits8(n) { return COUNT_BITS_TABLE[n & 0xff] } console.log('countOneBits32', countOneBits32(0b10101010000000001010101000000000)) console.log('countOneBits32', countOneBits32(0b10101011110000001010101000000000)) console.log('countOneBits16', countOneBits16(0b1010101000000000)) console.log('countOneBits8', countOneBits8(0b10000010))
int countBits(int x)
{
int n = 0;
if (x) do n++;
while(x=x&(x-1));
return n;
}
或者:
int countBits(int x) { return (x)? 1+countBits(x&(x-1)): 0; }
在我最初的回答7年半之后，@PeterMortensen质疑这是否是有效的C语法。我发布了一个在线编译器的链接，显示它实际上是完全有效的语法(代码如下)。
#include <stdio.h>
int countBits(int x)
{
int n = 0;
if (x) do n++; /* Totally Normal Valid code. */
while(x=x&(x-1)); /* Nothing to see here. */
return n;
}
int main(void) {
printf("%d\n", countBits(25));
return 0;
}
输出:
3
如果你想重新写清楚，它看起来是这样的:
if (x)
{
do
{
n++;
} while(x=x&(x-1));
}
但在我看来，这太过分了。
然而，我也意识到函数可以变得更短，但可能更神秘，写为:
int countBits(int x)
{
int n = 0;
while (x) x=(n++,x&(x-1));
return n;
}
对于那些想要在c++ 11中为任何无符号整数类型作为consexpr函数的人(tacklelib/include/tacklelib/utility/math.hpp):
#include <stdint.h>
#include <limits>
#include <type_traits>
const constexpr uint32_t uint32_max = (std::numeric_limits<uint32_t>::max)();
namespace detail
{
template <typename T>
inline constexpr T _count_bits_0(const T & v)
{
return v - ((v >> 1) & 0x55555555);
}
template <typename T>
inline constexpr T _count_bits_1(const T & v)
{
return (v & 0x33333333) + ((v >> 2) & 0x33333333);
}
template <typename T>
inline constexpr T _count_bits_2(const T & v)
{
return (v + (v >> 4)) & 0x0F0F0F0F;
}
template <typename T>
inline constexpr T _count_bits_3(const T & v)
{
return v + (v >> 8);
}
template <typename T>
inline constexpr T _count_bits_4(const T & v)
{
return v + (v >> 16);
}
template <typename T>
inline constexpr T _count_bits_5(const T & v)
{
return v & 0x0000003F;
}
template <typename T, bool greater_than_uint32>
struct _impl
{
static inline constexpr T _count_bits_with_shift(const T & v)
{
return
detail::_count_bits_5(
detail::_count_bits_4(
detail::_count_bits_3(
detail::_count_bits_2(
detail::_count_bits_1(
detail::_count_bits_0(v)))))) + count_bits(v >> 32);
}
};
template <typename T>
struct _impl<T, false>
{
static inline constexpr T _count_bits_with_shift(const T & v)
{
return 0;
}
};
}
template <typename T>
inline constexpr T count_bits(const T & v)
{
static_assert(std::is_integral<T>::value, "type T must be an integer");
static_assert(!std::is_signed<T>::value, "type T must be not signed");
return uint32_max >= v ?
detail::_count_bits_5(
detail::_count_bits_4(
detail::_count_bits_3(
detail::_count_bits_2(
detail::_count_bits_1(
detail::_count_bits_0(v)))))) :
detail::_impl<T, sizeof(uint32_t) < sizeof(v)>::_count_bits_with_shift(v);
}
谷歌测试库中的附加测试:
#include <stdlib.h>
#include <time.h>
namespace {
template <typename T>
inline uint32_t _test_count_bits(const T & v)
{
uint32_t count = 0;
T n = v;
while (n > 0) {
if (n % 2) {
count += 1;
}
n /= 2;
}
return count;
}
}
TEST(FunctionsTest, random_count_bits_uint32_100K)
{
srand(uint_t(time(NULL)));
for (uint32_t i = 0; i < 100000; i++) {
const uint32_t r = uint32_t(rand()) + (uint32_t(rand()) << 16);
ASSERT_EQ(_test_count_bits(r), count_bits(r));
}
}
TEST(FunctionsTest, random_count_bits_uint64_100K)
{
srand(uint_t(time(NULL)));
for (uint32_t i = 0; i < 100000; i++) {
const uint64_t r = uint64_t(rand()) + (uint64_t(rand()) << 16) + (uint64_t(rand()) << 32) + (uint64_t(rand()) << 48);
ASSERT_EQ(_test_count_bits(r), count_bits(r));
}
}
我给出了两个算法来回答这个问题，
package countSetBitsInAnInteger;
import java.util.Scanner;
public class UsingLoop {
public static void main(String[] args) {
Scanner in = new Scanner(System.in);
try {
System.out.println("Enter a integer number to check for set bits in it");
int n = in.nextInt();
System.out.println("Using while loop, we get the number of set bits as: " + usingLoop(n));
System.out.println("Using Brain Kernighan's Algorithm, we get the number of set bits as: " + usingBrainKernighan(n));
System.out.println("Using ");
}
finally {
in.close();
}
}
private static int usingBrainKernighan(int n) {
int count = 0;
while(n > 0) {
n& = (n-1);
count++;
}
return count;
}
/*
Analysis:
Time complexity = O(lgn)
Space complexity = O(1)
*/
private static int usingLoop(int n) {
int count = 0;
for(int i=0; i<32; i++) {
if((n&(1 << i)) != 0)
count++;
}
return count;
}
/*
Analysis:
Time Complexity = O(32) // Maybe the complexity is O(lgn)
Space Complexity = O(1)
*/
}
