[leetcode-191-Number of 1 Bits]

Write a function that takes an unsigned integer and returns the number of ’1‘ bits it has (also known as the Hamming weight).

For example, the 32-bit integer ’11‘ has binary representation 00000000000000000000000000001011, so the function should return 3.

思路1：向右移位，判断最后一位即可。

思路2：

 n & (n - 1) drops the lowest set bit. It‘s a neat little bit trick.

Let‘s use n = 00101100 as an example. This binary representation has three 1s.

If n = 00101100, then n - 1 = 00101011, so n & (n - 1) = 00101100 & 00101011 = 00101000. Count = 1.

If n = 00101000, then n - 1 = 00100111, so n & (n - 1) = 00101000 & 00100111 = 00100000. Count = 2.

If n = 00100000, then n - 1 = 00011111, so n & (n - 1) = 00100000 & 00011111 = 00000000. Count = 3.

n is now zero, so the while loop ends, and the final count (the numbers of set bits) is returned.

int hammingWeight(int n) 
{
  int weight = 0;
    while(n != 0)
    {
       if(n & 1 == 1)weight++;
       n>>=1;
    }
    return weight;        
 }
int hammingWeight(uint32_t n) {
    int count = 0;
    
    while (n) {
        n &= (n - 1);
        count++;
    }
    
    return count;
}

参考：

https://discuss.leetcode.com/topic/20120/c-solution-n-n-1

[leetcode-191-Number of 1 Bits]

原文：http://www.cnblogs.com/hellowooorld/p/6880578.html