标题：
通过率：	40.0%
难度：	简单

Write a function that takes an unsigned integer and returns the number of ’1' bits it has (also known as the ).

For example, the 32-bit integer ’11' has binary representation 00000000000000000000000000001011, so the function should return 3.

Credits:

Special thanks to  for adding this problem and creating all test cases.

 

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本题就是数给的十进制数的二进制数有几个1。

最基本的做法就是32位扫描一遍进行查，我查看网页后找到一种简单的方法，减1法，每次把二进制减一，一直减到0为止，

n=n&(n-1)；

具体代码如下：

1 public class Solution { 2     // you need to treat n as an unsigned value 3     public int hammingWeight(int n) { 4         int res=0; 5         while(n!=0){ 6             n=n&(n-1); 7             res++; 8         } 9         return res;10     }11 }