191. 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.

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

解题思路是n在右移的过程中，判断最低位是不是1，是1统计加1，不是1，不做任何事。代码如下：

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

这里注意，在java里，>>是数学右移，最该为的补位会跟符号位相同，那么如果是2147483648，它代表 -2147483648，100000000，永远也得不到0。而>>>是逻辑右移，不关心符号位，永远用0补位，这是我们这道题需要的。

另一种思路是 n = n & (n - 1)，直到n = 0为止。可以不用管标识位，代码如下：

int hammingWeight(uint32_t n) {    int count = 0;        while (n) {        n &= (n - 1);        count++;    }        return count;}