# 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.


题目翻译:
给出一个整数,求它包含二进制1的位数。例如,32位整数`11`的二进制表达形式是`00000000000000000000000000001011`,那么函数应该返回3。

题目分析:
设输入的数为n,把n与1做二进制的与(AND)运算,即可判断它的最低位是否为1。如果是的话,把计数变量加一。然后把n向右移动一位,重复上述操作。当n变为0时,终止算法,输出结果。

时间复杂度:O(n)
空间复杂度:O(1)

代码如下:

```c++
class Solution {
public:
int hammingWeight(uint32_t n) {
int count = 0;
while (n > 0) {
count += n & 1;
n >>= 1;
}
return count;
}
};
```