121_Number_of_1_bits

// https://leetcode.com/problems/number-of-1-bits/

public class Solution {
    // you need to treat n as an unsigned value
    public int hammingWeight(int n) {
        int bits = 0;
        int mask = 1;
        for(int i=0;i<32;i++){
            if((n&mask)!=0)
                bits++;
            mask<<=1;
        }
        return bits;
    }
}

/*
We can make the previous algorithm simpler and a little faster. Instead of checking every bit of the number, we repeatedly flip the least-significant 11-bit of the number to 00, and add 11 to the sum. As soon as the number becomes 00, we know that it does not have any more 11-bits, and we return the sum.

The key idea here is to realize that for any number nn, doing a bit-wise AND of nn and n - 1n−1 flips the least-significant 11-bit in nn to 00. Why? Consider the binary representations of nn and n - 1n−1.
In the binary representation, the least significant 11-bit in nn always corresponds to a 00-bit in n - 1n−1. Therefore, anding the two numbers nn and n - 1n−1 always flips the least significant 11-bit in nn to 00, and keeps all other bits the same.
*/

public int hammingWeight(int n) {
    int sum = 0;
    while (n != 0) {
        sum++;
        n &= (n - 1);
    }
    return sum;
}