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.

Credits:
Special thanks to @ts for adding this problem and creating all test cases.

参考编程之美120页

方法1：使用位操作

代码如下：

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

运行结果：时间复杂度为O(logV),即二进制位数。

方法二：n&=(n-1)

代码如下：

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

运行结果：