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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:

(easy)LeetCode 191.Number of 1 Bits
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原文地址:http://www.cnblogs.com/mlz-2019/p/4679860.html