以整数计数设置位

编写一个高效的程序，以整数的二进制表示形式计算1的数量。
例子 ：

Input : n = 6
Output : 2
Binary representation of 6 is 110 and has 2 set bits

Input : n = 13
Output : 3
Binary representation of 13 is 1101 and has 3 set bits

1.简单方法遍历整数中的所有位，检查是否设置了位，然后再递增设置的位计数。参见下面的程序。

C++

// C++ program to Count set
// bits in an integer
#include 
using namespace std;
 
/* Function to get no of set bits in binary
representation of positive integer n */
unsigned int countSetBits(unsigned int n)
{
    unsigned int count = 0;
    while (n) {
        count += n & 1;
        n >>= 1;
    }
    return count;
}
 
/* Program to test function countSetBits */
int main()
{
    int i = 9;
    cout << countSetBits(i);
    return 0;
}
 
// This code is contributed
// by Akanksha Rai

C

// C program to Count set
// bits in an integer
#include 
 
/* Function to get no of set bits in binary
   representation of positive integer n */
unsigned int countSetBits(unsigned int n)
{
    unsigned int count = 0;
    while (n) {
        count += n & 1;
        n >>= 1;
    }
    return count;
}
 
/* Program to test function countSetBits */
int main()
{
    int i = 9;
    printf("%d", countSetBits(i));
    return 0;
}

Java

// Java program to Count set
// bits in an integer
import java.io.*;
 
class countSetBits {
    /* Function to get no of set
    bits in binary representation
    of positive integer n */
    static int countSetBits(int n)
    {
        int count = 0;
        while (n > 0) {
            count += n & 1;
            n >>= 1;
        }
        return count;
    }
 
    // driver program
    public static void main(String args[])
    {
        int i = 9;
        System.out.println(countSetBits(i));
    }
}
 
// This code is contributed by Anshika Goyal.

Python3

# Python3 program to Count set
# bits in an integer
 
# Function to get no of set bits in binary
# representation of positive integer n */
def  countSetBits(n):
    count = 0
    while (n):
        count += n & 1
        n >>= 1
    return count
 
 
# Program to test function countSetBits */
i = 9
print(countSetBits(i))
 
# This code is contributed by
# Smitha Dinesh Semwal

C#

// C# program to Count set
// bits in an integer
using System;
 
class GFG {
    // Function to get no of set
    // bits in binary representation
    // of positive integer n
    static int countSetBits(int n)
    {
        int count = 0;
        while (n > 0) {
            count += n & 1;
            n >>= 1;
        }
        return count;
    }
 
    // Driver Code
    public static void Main()
    {
        int i = 9;
        Console.Write(countSetBits(i));
    }
}
 
// This code is contributed by Sam007

PHP

>= 1;
    }
    return $count;
}
 
// Driver Code
$i = 9;
echo countSetBits($i);
 
// This code is contributed by ajit
?>

Javascript

    // Javascript program to Count set
    // bits in an integer
 
    /* Function to get no of set bits in binary
    representation of positive integer n */
    function countSetBits(n)
    {
      var count = 0;
      while (n)
      {
        count += n & 1;
        n >>= 1;
      }
      return count;
    }
 
    /* Program to test function countSetBits */
    var i = 9;
    document.write(countSetBits(i));
 
// This code is contributed by noob2000.
  

输出 ：

2

时间复杂度： (-)(logn)(logn的Theta)
递归方法：

C++

// cpp implementation of recursive
// approach to find the number
// of set bits in binary representation
// of positive integer n
#include 
using namespace std;
 
// recursive function to count set bits
int countSetBits(int n)
{
    // base case
    if (n == 0)
        return 0;
 
    else
 
        // if last bit set add 1 else add 0
        return (n & 1) + countSetBits(n >> 1);
}
 
// driver code
int main()
{
    // get value from user
    int n = 9;
 
    // function calling
    cout << countSetBits(n);
 
    return 0;
}
 
// This code is contributed by Raj.

Java

// Java implementation of recursive
// approach to find the number
// of set bits in binary representation
// of positive integer n
import java.io.*;
 
class GFG {
 
    // recursive function to count set bits
    public static int countSetBits(int n)
    {
 
        // base case
        if (n == 0)
            return 0;
 
        else
 
            // if last bit set add 1 else add 0
            return (n & 1) + countSetBits(n >> 1);
    }
 
    // Driver code
    public static void main(String[] args)
    {
 
        // get value from user
        int n = 9;
 
        // function calling
        System.out.println(countSetBits(n));
    }
}
 
// This code is contributes by sunnysingh

Python3

# Python3 implementation of recursive
# approach to find the number of set
# bits in binary representation of
# positive integer n
 
def countSetBits( n):
     
    # base case
    if (n == 0):
        return 0
 
    else:
 
        # if last bit set add 1 else
        # add 0
        return (n & 1) + countSetBits(n >> 1)
         
# Get value from user
n = 9
 
# Function calling
print( countSetBits(n))    
         
# This code is contributed by sunnysingh

C＃

// C# implementation of recursive
// approach to find the number of
// set bits in binary representation
// of positive integer n
using System;
 
class GFG {
 
    // recursive function
    // to count set bits
    public static int countSetBits(int n)
    {
 
        // base case
        if (n == 0)
            return 0;
 
        else
 
            // if last bit set
            // add 1 else add 0
            return (n & 1) + countSetBits(n >> 1);
    }
 
    // Driver code
    static public void Main()
    {
 
        // get value
        // from user
        int n = 9;
 
        // function calling
        Console.WriteLine(countSetBits(n));
    }
}
 
// This code is contributed by aj_36

的PHP

> 1);
}
 
// Driver code
 
// get value from user
$n = 9;
 
// function calling
echo countSetBits($n);
 
// This code is contributed by m_kit.
?>

Java脚本

 
// Javascript implementation of recursive
// approach to find the number
// of set bits in binary representation
// of positive integer n
 
// recursive function to count set bits
function countSetBits(n)
{
 
    // base case
    if (n == 0)
        return 0;
 
    else
 
        // if last bit set add 1 else add 0
        return (n & 1) + countSetBits(n >> 1);
}
 
// driver code
 
    // get value from user
    let n = 9;
 
    // function calling
    document.write(countSetBits(n));
 
 
// This code is contributed by Mayank Tyagi

输出 ：

2

2. Brian Kernighan的算法：
从十进制数中减去1将翻转最右边的设置位(即1)之后的所有位，包括最右边的设置位。
例如 ：
10的二进位是00001010
二进制9是00001001
8的二进位是00001000
二进制中的7是00000111
因此，如果我们将数字减1并对其本身按位与(n＆(n-1))，则会取消设置最右边的位。如果我们在循环中执行n＆(n-1)并计算循环执行的次数，我们将获得设置的位数。
该解决方案的优点在于，它循环的次数等于给定整数中的设置位数。

1  Initialize count: = 0
   2  If integer n is not zero
      (a) Do bitwise & with (n-1) and assign the value back to n
          n: = n&(n-1)
      (b) Increment count by 1
      (c) go to step 2
   3  Else return count

Brian Kernighan算法的实现：

C++

// C++ program to Count set
// bits in an integer
#include 
using namespace std;
class gfg {
    /* Function to get no of set bits in binary
representation of passed binary no. */
public:
    unsigned int countSetBits(int n)
    {
        unsigned int count = 0;
        while (n) {
            n &= (n - 1);
            count++;
        }
        return count;
    }
};
/* Program to test function countSetBits */
int main()
{
    gfg g;
    int i = 9;
    cout << g.countSetBits(i);
    return 0;
}

C

// C program to Count set
// bits in an integer
#include 
 
/* Function to get no of set bits in binary
   representation of passed binary no. */
unsigned int countSetBits(int n)
{
    unsigned int count = 0;
    while (n) {
        n &= (n - 1);
        count++;
    }
    return count;
}
 
/* Program to test function countSetBits */
int main()
{
    int i = 9;
    printf("%d", countSetBits(i));
    getchar();
    return 0;
}

Java

// Java program to Count set
// bits in an integer
import java.io.*;
 
class countSetBits {
    /* Function to get no of set
    bits in binary representation
    of passed binary no. */
    static int countSetBits(int n)
    {
        int count = 0;
        while (n > 0) {
            n &= (n - 1);
            count++;
        }
        return count;
    }
 
    // driver program
    public static void main(String args[])
    {
        int i = 9;
        System.out.println(countSetBits(i));
    }
}
 
// This code is contributed by Anshika Goyal.

Python3

# Function to get no of set bits in binary
# representation of passed binary no. */
def countSetBits(n):
 
    count = 0
    while (n):
        n &= (n-1)
        count+= 1
     
    return count
 
 
# Program to test function countSetBits
i = 9
print(countSetBits(i))
  
# This code is contributed by
# Smitha Dinesh Semwal

C＃

// C# program to Count set
// bits in an integer
using System;
 
class GFG {
 
    /* Function to get no of set
    bits in binary representation
    of passed binary no. */
    static int countSetBits(int n)
    {
        int count = 0;
        while (n > 0) {
            n &= (n - 1);
            count++;
        }
        return count;
    }
 
    // Driver Code
    static public void Main()
    {
        int i = 9;
        Console.WriteLine(countSetBits(i));
    }
}
 
// This code is contributed by ajit

Java脚本

 
// JavaScript program to Count set
// bits in an integerclass
 
/* Function to get no of set
bits in binary representation
of passed binary no. */
 
function countSetBits(n)
{
    var count = 0;
    while (n > 0) {
        n &= (n - 1);
        count++;
    }
    return count;
}
 
// driver program
var i = 9;
document.write(countSetBits(i));
 
// This code is contributed by 29AjayKumar
 

输出 ：

2

Brian Kernighan算法的示例：

n =  9 (1001)
   count = 0

   Since 9 > 0, subtract by 1 and do bitwise & with (9-1)
   n = 9&8  (1001 & 1000)
   n = 8
   count  = 1

   Since 8 > 0, subtract by 1 and do bitwise & with (8-1)
   n = 8&7  (1000 & 0111)
   n = 0
   count = 2

   Since n = 0, return count which is 2 now.

时间复杂度： O(logn)
递归方法：

C++

// CPP implementation for recursive
// approach to find the number of set
// bits using Brian Kernighan’s Algorithm
#include 
using namespace std;
 
// recursive function to count set bits
int countSetBits(int n)
{
    // base case
    if (n == 0)
        return 0;
    else
        return 1 + countSetBits(n & (n - 1));
}
 
// driver code
int main()
{
    // get value from user
    int n = 9;
 
    // function calling
    cout << countSetBits(n);
 
    return 0;
}
 
// This code is contributed by Raj.

Java

// Java implementation for recursive
// approach to find the number of set
// bits using Brian Kernighan Algorithm
import java.io.*;
 
class GFG {
 
    // recursive function to count set bits
    public static int countSetBits(int n)
    {
 
        // base case
        if (n == 0)
            return 0;
        else
            return 1 + countSetBits(n & (n - 1));
    }
 
    // Driver function
    public static void main(String[] args)
    {
 
        // get value from user
        int n = 9;
 
        // function calling
        System.out.println(countSetBits(n));
    }
}
 
// This code is contributed by sunnysingh

Python3

# Python3 implementation for
# recursive approach to find
# the number of set bits using
# Brian Kernighan’s Algorithm
 
# recursive function to count
# set bits
def countSetBits(n):
 
    # base case
    if (n == 0):
        return 0
    else:
        return 1 + countSetBits(n & (n - 1))
             
             
# Get value from user
n = 9
     
# function calling
print(countSetBits(n))
 
# This code is contributed by sunnysingh

C＃

// C# implementation for recursive
// approach to find the number of set
// bits using Brian Kernighan Algorithm
using System;
 
class GFG {
 
    // recursive function
    // to count set bits
    public static int countSetBits(int n)
    {
 
        // base case
        if (n == 0)
            return 0;
        else
            return 1 + countSetBits(n & (n - 1));
    }
 
    // Driver Code
    static public void Main()
    {
 
        // get value from user
        int n = 9;
 
        // function calling
        Console.WriteLine(countSetBits(n));
    }
}
 
// This code is contributed by aj_36

Java脚本

 
// Javascript implementation for recursive
// approach to find the number of set
// bits using Brian Kernighan’s Algorithm
 
// recursive function to count set bits
function countSetBits(n)
{
    // base case
    if (n == 0)
        return 0;
    else
        return 1 + countSetBits(n & (n - 1));
}
 
// driver code
// get value from user
 
var n = 9;
// function calling
 
document.write(countSetBits(n));
 

输出 ：

2

3.使用查找表：我们可以使用查找表对O(1)时间中的位进行计数。有关详细信息，请参见http://graphics.stanford.edu/~seander/bithacks.html#CountBitsSetTable。
下面是上述方法的实现：

C++

// C++ implementation of the approach
#include 
using namespace std;
 
int BitsSetTable256[256];
 
// Function to initialise the lookup table
void initialize()
{
 
    // To initially generate the
    // table algorithmically
    BitsSetTable256[0] = 0;
    for (int i = 0; i < 256; i++)
    {
        BitsSetTable256[i] = (i & 1) +
        BitsSetTable256[i / 2];
    }
}
 
// Function to return the count
// of set bits in n
int countSetBits(int n)
{
    return (BitsSetTable256[n & 0xff] +
            BitsSetTable256[(n >> 8) & 0xff] +
            BitsSetTable256[(n >> 16) & 0xff] +
            BitsSetTable256[n >> 24]);
}
 
// Driver code
int main()
{
    // Initialise the lookup table
    initialize();
    int n = 9;
    cout << countSetBits(n);
}
 
// This code is contributed by Sanjit_Kumar

Java

// Java implementation of the approach
class GFG {
 
    // Lookup table
    static int[] BitsSetTable256 = new int[256];
 
    // Function to initialise the lookup table
    public static void initialize()
    {
 
        // To initially generate the
        // table algorithmically
        BitsSetTable256[0] = 0;
        for (int i = 0; i < 256; i++) {
            BitsSetTable256[i] = (i & 1) + BitsSetTable256[i / 2];
        }
    }
 
    // Function to return the count
    // of set bits in n
    public static int countSetBits(int n)
    {
        return (BitsSetTable256[n & 0xff]
                + BitsSetTable256[(n >> 8) & 0xff]
                + BitsSetTable256[(n >> 16) & 0xff]
                + BitsSetTable256[n >> 24]);
    }
 
    // Driver code
    public static void main(String[] args)
    {
 
        // Initialise the lookup table
        initialize();
        int n = 9;
        System.out.print(countSetBits(n));
    }
}

Python

# Python implementation of the approach
BitsSetTable256 = [0] * 256
 
# Function to initialise the lookup table
def initialize():
     
    # To initially generate the
    # table algorithmically
    BitsSetTable256[0] = 0
    for i in range(256):
        BitsSetTable256[i] = (i & 1) + BitsSetTable256[i // 2]
 
# Function to return the count
# of set bits in n
def countSetBits(n):
    return (BitsSetTable256[n & 0xff] +
            BitsSetTable256[(n >> 8) & 0xff] +
            BitsSetTable256[(n >> 16) & 0xff] +
            BitsSetTable256[n >> 24])
 
# Driver code
 
# Initialise the lookup table
initialize()
n = 9
print(countSetBits(n))
 
# This code is contributed by SHUBHAMSINGH10

C＃

// C# implementation of the approach
using System;
using System.Collections.Generic;
 
class GFG
{
 
    // Lookup table
    static int[] BitsSetTable256 = new int[256];
 
    // Function to initialise the lookup table
    public static void initialize()
    {
 
        // To initially generate the
        // table algorithmically
        BitsSetTable256[0] = 0;
        for (int i = 0; i < 256; i++)
        {
            BitsSetTable256[i] = (i & 1) + BitsSetTable256[i / 2];
        }
    }
 
    // Function to return the count
    // of set bits in n
    public static int countSetBits(int n)
    {
        return (BitsSetTable256[n & 0xff]
                + BitsSetTable256[(n >> 8) & 0xff]
                + BitsSetTable256[(n >> 16) & 0xff]
                + BitsSetTable256[n >> 24]);
    }
 
    // Driver code
    public static void Main(String[] args)
    {
 
        // Initialise the lookup table
        initialize();
        int n = 9;
        Console.Write(countSetBits(n));
    }
}
 
// This code is contributed by 29AjayKumar

Java脚本

 
// javascript implementation of the approach  
 
var BitsSetTable256 = Array.from({length: 256}, (_, i) => 0);
 
// Function to initialise the lookup table
function initialize()
{
 
    // To initially generate the
    // table algorithmically
    BitsSetTable256[0] = 0;
    for (var i = 0; i < 256; i++) {
        BitsSetTable256[i] = (i & 1) +
        BitsSetTable256[parseInt(i / 2)];
    }
}
 
// Function to return the count
// of set bits in n
function countSetBits(n)
{
    return (BitsSetTable256[n & 0xff]
            + BitsSetTable256[(n >> 8) & 0xff]
            + BitsSetTable256[(n >> 16) & 0xff]
            + BitsSetTable256[n >> 24]);
}
 
// Driver code
 // Initialise the lookup table
initialize();
var n = 9;
document.write(countSetBits(n));
 
// This code is contributed by 29AjayKumar
 

输出：

2

我们可以在计数要翻转以将A转换为B的位数时找到计数设置位的一种用法
注意：在GCC中，我们可以使用__builtin_popcount()直接计数设置位。因此，我们可以避免使用单独的函数来计数设置位。

C++

// C++ program to demonstrate __builtin_popcount()
#include 
using namespace std;
 
int main()
{
    cout << __builtin_popcount(4) << endl;
    cout << __builtin_popcount(15);
 
    return 0;
}

Java

// java program to demonstrate
// __builtin_popcount()
 
import java.io.*;
 
class GFG {
 
    // Driver code
    public static void main(String[] args)
    {
 
        System.out.println(Integer.bitCount(4));
        System.out.println(Integer.bitCount(15));
    }
}
 
// This code is contributed by Raj

Python3

# Python3 program to demonstrate __builtin_popcount()
 
print(bin(4).count('1'));
print(bin(15).count('1'));
 
# This code is Contributed by mits

C＃

// C# program to demonstrate
// __builtin_popcount()
using System;
using System.Linq;
 
class GFG {
 
    // Driver code
    public static void Main()
    {
 
        Console.WriteLine(Convert.ToString(4, 2).Count(c = > c == '1'));
        Console.WriteLine(Convert.ToString(15, 2).Count(c = > c == '1'));
    }
}
 
// This code is contributed by mits

输出 ：

1
4

4.用该位映射数字。它只是维护数字到位的映射(或数组)，以进行半字节。半字节包含4位。因此，我们需要最多15个数组。
int num_to_bits [16] = {0，1，1，2，1，2，2，2，3，1，2，2，3，2，3，3，4};
现在我们只需要递归地得到给定的long / int / word等字节。

C++

// C++ program to count set bits by pre-storing
// count set bits in nibbles.
#include 
using namespace std;
 
int num_to_bits[16] = { 0, 1, 1, 2, 1, 2, 2, 3,
                        1, 2, 2, 3, 2, 3, 3, 4 };
 
/* Recursively get nibble of a given number
and map them in the array */
unsigned int countSetBitsRec(unsigned int num)
{
    int nibble = 0;
    if (0 == num)
        return num_to_bits[0];
 
    // Find last nibble
    nibble = num & 0xf;
 
    // Use pre-stored values to find count
    // in last nibble plus recursively add
    // remaining nibbles.
    return num_to_bits[nibble] + countSetBitsRec(num >> 4);
}
 
// Driver code
int main()
{
    int num = 31;
    cout << countSetBitsRec(num);
    return 0;
}
 
// This code is contributed by rathbhupendra

C

// C program to count set bits by pre-storing
// count set bits in nibbles.
#include 
 
int num_to_bits[16] = { 0, 1, 1, 2, 1, 2, 2, 3,
                        1, 2, 2, 3, 2, 3, 3, 4 };
 
/* Recursively get nibble of a given number
  and map them in the array  */
unsigned int countSetBitsRec(unsigned int num)
{
    int nibble = 0;
    if (0 == num)
        return num_to_bits[0];
 
    // Find last nibble
    nibble = num & 0xf;
 
    // Use pre-stored values to find count
    // in last nibble plus recursively add
    // remaining nibbles.
    return num_to_bits[nibble] + countSetBitsRec(num >> 4);
}
 
// Driver code
int main()
{
    int num = 31;
    printf("%d\n", countSetBitsRec(num));
}

Java

// Java program to count set bits by pre-storing
// count set bits in nibbles.
 
class GFG {
    static int[] num_to_bits = new int[] { 0, 1, 1, 2, 1, 2, 2,
                                           3, 1, 2, 2, 3, 2, 3, 3, 4 };
 
    /* Recursively get nibble of a given number
and map them in the array */
    static int countSetBitsRec(int num)
    {
        int nibble = 0;
        if (0 == num)
            return num_to_bits[0];
 
        // Find last nibble
        nibble = num & 0xf;
 
        // Use pre-stored values to find count
        // in last nibble plus recursively add
        // remaining nibbles.
        return num_to_bits[nibble] + countSetBitsRec(num >> 4);
    }
 
    // Driver code
    public static void main(String[] args)
    {
        int num = 31;
        System.out.println(countSetBitsRec(num));
    }
}
// this code is contributed by mits

Python3

# Python3 program to count set bits by pre-storing
# count set bits in nibbles.
 
num_to_bits =[0, 1, 1, 2, 1, 2, 2, 3, 1, 2, 2, 3, 2, 3, 3, 4];
 
# Recursively get nibble of a given number
# and map them in the array
def countSetBitsRec(num):
    nibble = 0;
    if(0 == num):
        return num_to_bits[0];
     
    # Find last nibble
    nibble = num & 0xf;
     
    # Use pre-stored values to find count
    # in last nibble plus recursively add
    # remaining nibbles.
     
    return num_to_bits[nibble] + countSetBitsRec(num >> 4);
  
 
# Driver code
  
num = 31;
print(countSetBitsRec(num));
 
 
# this code is contributed by mits

C＃

// C# program to count set bits by pre-storing
// count set bits in nibbles.
 
class GFG {
    static int[] num_to_bits = new int[16] { 0, 1, 1, 2, 1, 2, 2,
                                             3, 1, 2, 2, 3, 2, 3, 3, 4 };
 
    /* Recursively get nibble of a given number
and map them in the array */
    static int countSetBitsRec(int num)
    {
        int nibble = 0;
        if (0 == num)
            return num_to_bits[0];
 
        // Find last nibble
        nibble = num & 0xf;
 
        // Use pre-stored values to find count
        // in last nibble plus recursively add
        // remaining nibbles.
        return num_to_bits[nibble] + countSetBitsRec(num >> 4);
    }
 
    // Driver code
    static void Main()
    {
        int num = 31;
        System.Console.WriteLine(countSetBitsRec(num));
    }
}
// this code is contributed by mits

的PHP

> 4);
}
 
// Driver code
$num = 31;
echo (countSetBitsRec($num));
 
// This code is contributed by mits
?>

输出 ：

5

时间复杂度： O(1)
存储复杂度： O(1)无论给定的数字是short，int，long还是long long，我们只需要16个大小恒定的数组即可。

5.检查数字中的每一位：

检查数字中的每个位是否已设置。该数字是按位与的，且具有2的幂，因此，如果结果不等于0，我们就会知道该位置中的特定位已设置。

C

#include 
 
// Check each bit in a number is set or not
// and return the total count of the set bits.
int countSetBits(int N)
{
    int count = 0;
   
    // (1 << i) = pow(2, i)
    for (int i = 0; i < sizeof(int) * 8; i++) {
        if (N & (1 << i))
            count++;
    }
    return count;
}
 
// Driver Code
int main()
{
    int N = 15;
 
    printf("%d", countSetBits(N));
    return 0;
}

C++

#include 
using namespace std;
 
// Check each bit in a number is set or not
// and return the total count of the set bits.
int countSetBits(int N)
{
    int count = 0;
    // (1 << i) = pow(2, i)
    for (int i = 0; i < sizeof(int) * 8; i++) {
        if (N & (1 << i))
            count++;
    }
    return count;
}
 
int main()
{
 
    int N = 15;
 
    cout << countSetBits(N) << endl;
    return 0;
}

Java

public class GFG
{
   
  // Check each bit in a number is set or not
  // and return the total count of the set bits.
  static int countSetBits(int N)
  {
    int count = 0;
    // (1 << i) = pow(2, i)
    for (int i = 0; i < 4 * 8; i++)
    {
      if ((N & (1 << i)) != 0)
        count++;
    }
    return count;
  }
 
  // Driver code
  public static void main(String[] args)
  {
    int N = 15;
    System.out.println(countSetBits(N));
  }
}
 
// This code is contributed by divyeshrabadiya07.

Python3

# Check each bit in a number is set or not
# and return the total count of the set bits
def countSetBits(N):
  count = 0
 
  # (1 << i) = pow(2, i)
  for i in range(4*8):
    if(N & (1 << i)):
      count += 1
      return count
 
    # Driver code
    N = 15
    print(countSetBits(N))
 
    # This code is contributed by avanitrachhadiya2155

C＃

using System;
class GFG
{
 
  // Check each bit in a number is set or not
  // and return the total count of the set bits.
  static int countSetBits(int N)
  {
    int count = 0;
 
    // (1 << i) = pow(2, i)
    for (int i = 0; i < 4 * 8; i++)
    {
      if ((N & (1 << i)) != 0)
        count++;
    }
    return count;
  }
 
  // Driver code
  static void Main()
  {
    int N = 15;
    Console.WriteLine(countSetBits(N));
  }
}
 
// This code is contributed by divyesh072019.

Java脚本

   
  // Check each bit in a number is set or not
  // and return the total count of the set bits.
  function countSetBits(N)
  {
    var count = 0;
    // (1 << i) = pow(2, i)
    for (i = 0; i < 4 * 8; i++)
    {
      if ((N & (1 << i)) != 0)
        count++;
    }
    return count;
  }
 
  // Driver code
  var N = 15;
  document.write(countSetBits(N));
 
// This code is contributed by Amit Katiyar

输出

4

使用查找表对整数中的设置位进行计数

提问人：Adobe，Brocade，Cisco，Juniper Networks，Qualcomm

参考：
http://graphics.stanford.edu/~seander/bithacks.html#CountBitsSetNaive