###### tags: `Leetcode` `easy` `bit` `python` `c++`

# 191. Number of 1 Bits

## [題目連結:] https://leetcode.com/problems/number-of-1-bits/

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

Note:

* Note that in some languages, such as Java, there is no unsigned integer type. In this case, the input will be given as a signed integer type. It should not affect your implementation, as the integer's internal binary representation is the same, whether it is signed or unsigned.
* In Java, the compiler represents the signed integers using 2's complement notation. Therefore, in Example 3, the input represents the signed integer. -3.


#### [圖片來源:] https://leetcode.com/problems/number-of-1-bits/

## 解題想法:

兄弟題目: [190. Reverse Bits](/rGFkEUTqTCmcJmMfmRiHwA)

取最後一位: 某數&1

## Python:

class Solution(object):
    def hammingWeight(self, n):
        """
        :type n: int
        :rtype: int
        """
        ans = 0
        while n:
            #若n 最後一碼為0 則不會進入迴圈
            if (n&1) :
                ans = ans +1
            #n向右移
            n = n >> 1
        return ans

if __name__ == '__main__':
    result = Solution()
    n = int('00000000000000000000000000001011',2)
    ans = result.hammingWeight(n)
    print(n,ans)

## C++:

 #include<iostream>
 using namespace std;

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

int main(){
    Solution res;
    // 0bxxxxxx  binary string representation
    uint32_t n= 0b00000000000000000000000000001011;
    int ans=res.hammingWeight(n);
    cout<<ans<<endl;
    return 0;
}