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.

public class Solution {
    // you need to treat n as an unsigned value
    public int hammingWeight(int n) {
        n = n - ((n >>> 1) & 0x55555555);
        n = (n & 0x33333333) + ((n >>> 2) & 0x33333333);
        n = (n + (n >>> 4)) & 0x0f0f0f0f;
        n = n + (n >>> 8);
        n = n + (n >>> 16);
        return n & 0x3f;
    }
}

public class Solution {
    // you need to treat n as an unsigned value
    public int hammingWeight(int n) {
        int result = 0;
        while(n > 1){
            result += n % 2;
            n = (int)(n / 2);
        }
        if(n == 1) result ++;
        return result;
    }
}

it fail at n == 2147483648：

Because 2147483648 is not greater than 2147483648, and then shift to left 1 bit make bit == 0 instead of 2147483648 * 2. Then it fall into the infinity loop where bit always be 0.

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