LeetCode 191. Number of 1 Bits (Easy)

题目描述：

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

Example：

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

题目大意：找出一个无符号整数的二进制形式的1的个数。

思路：直接做也不是不可以（循环找1）。但是我从网上看到了一个很巧妙的方法：当n不等于0时，n = n & (n - 1)，循环次数就是1的个数。思考了一下：n & n - 1可以令最低位的1消掉，所以每一次消除一个最低位的1，消除的次数就是1的个数了。

以题目为例：

n	n-1	n & n - 1
00000000000000000000000000001011	00000000000000000000000000001010	00000000000000000000000000001010
00000000000000000000000000001010	00000000000000000000000000001001	00000000000000000000000000001000
00000000000000000000000000001000	00000000000000000000000000000111	00000000000000000000000000000000

c++代码：

class Solution {
public:
int hammingWeight(uint32_t n) {
int ans = 0;
while (n)
{
n = n & (n - 1);
ans++;
}
return ans;
}
};