ama-non

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

Example 1:

Input: 00000000000000000000000000001011

Output: 3

Explanation: The input binary string 00000000000000000000000000001011 has a total of three '1' bits.

Example 2:

Input: 00000000000000000000000010000000

Output: 1

Explanation: The input binary string 00000000000000000000000010000000 has a total of one '1' bit.

Example 3:

Input: 11111111111111111111111111111101

Output: 31

Explanation: The input binary string 11111111111111111111111111111101 has a total of thirty one '1' bits.

Note:

• Note that in some languages such as Java, there is no unsigned integer type. In this case, the input will be given as signed integer type and should not affect your implementation, as the internal binary representation of the integer is the same whether it is signed or unsigned.

Follow up:

If this function is called many times, how would you optimize it?

思路一：

最直接的方法，做32次右移，统计出1的个数。

class Solution {
public:
    int hammingWeight(uint32_t n) {
        int bits = 0;
        uint32_t mask = 1;
        for (int i = 0; i < 32; i++) {
            if (n & mask) 
                bits++;
            mask <<= 1;
        }
        return bits;
    }
};

时间复杂度o(1),空间复杂度o(1)

思路二：

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

时间复杂度o(1),空间复杂度o(1)

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.