191. Number of 1 Bits

Iterate

public class Solution {
    // you need to treat n as an unsigned value
    public int hammingWeight(int n) {
        int weight = 0;
        while (n != 0) {
            if ((n & 1) == 1) {
                weight++;
            }
            n >>>= 1;
        }

        return weight;
    }
}

n & (n − 1)

public class Solution {
    // you need to treat n as an unsigned value
    public int hammingWeight(int n) {
        int weight = 0;
        while (n != 0) { // 注意 i 可能为负数，所以此处需要使用不等于 0 判断，而不能使用大于 0 进行判断
            weight++;
            n &= n - 1;
        }

        return weight;
    }
}

Reference

191. Number of 1 Bits
剑指 Offer 15. 二进制中1的个数