Count set bits in an integer

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本文介绍了一种高效算法，用于计算整数二进制表示中1的个数。采用Brian Kernighan算法，通过循环减一并进行按位与操作来消除最右侧的1位，直至所有位被清除。

reference:

http://www.geeksforgeeks.org/count-set-bits-in-an-integer/

Problem Definition:

Write an efficient program to count number of 1s in binary representation of an integer.

Solution:

Brian Kernighan’s Algorithm:

Subtraction of 1 from a number toggles all the bits (from right to left) till the rightmost set bit(including the righmost set bit). Soif we subtract a number by 1 and do bitwise & with itself (n & (n-1)), we unset the righmost set bit. If we do n & (n-1) in a loop and count the no of times loop executes we get the set bit count.

 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

Code:

/* Function to get no of set bits in binary
   representation of passed binary no. */
int countSetBits(int n)
{
    unsigned int count = 0;
    while (n)
    {
      n &= (n-1) ;
      count++;
    }
    return count;
}