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.

Solution 1: Naive way: use mask to check every bit of the number.

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

Solution 2: line 7 uses n&n-1 to repeatedly flip the least-significant 1-bit of the number to 0 and add 1 to the result, until the number becomes 0. So we don't need to check every bit of the number.

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

Solution 3: https://en.wikipedia.org/wiki/Hamming_weight

class Solution {
public:
    int hammingWeight(uint32_t n) {
        n = (n & 0x55555555) + ((n >> 1) & 0x55555555);
        n = (n & 0x33333333) + ((n >> 2) & 0x33333333);
        n = (n & 0x0f0f0f0f) + ((n >> 4) & 0x0f0f0f0f);
        n = (n & 0x00ff00ff) + ((n >> 8) & 0x00ff00ff);
        n = (n & 0x0000ffff) + ((n >> 16) & 0x0000ffff);
        return (n);
    }
};