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

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;}

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