题目

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

Note:

Note that in some languages such as Java, there is no unsigned integer type. In this case, the input will be given as a signed integer type. It should not affect your implementation, as the integer’s internal binary representation is the same, whether it is signed or unsigned.
In Java, the compiler represents the signed integers using 2’s complement notation. Therefore, in Example 3 above, the input represents the signed integer. -3.
Follow up: If this function is called many times, how would you optimize it?

Example 1:
Input: n = 00000000000000000000000000001011
Output: 3
Explanation: The input binary string 00000000000000000000000000001011 has a total of three ‘1’ bits.

Example 2:
Input: n = 00000000000000000000000010000000
Output: 1
Explanation: The input binary string 00000000000000000000000010000000 has a total of one ‘1’ bit.

Example 3:
Input: n = 11111111111111111111111111111101
Output: 31
Explanation: The input binary string 11111111111111111111111111111101 has a total of thirty one ‘1’ bits.

Constraints:

The input must be a binary string of length 32

解析

输入为固定长的32bits, 所以咋做都不会超时.
有一种$O(\log{2}{N}$的方法. 其本质上是一种分治法:

先将每1bit与相邻与相邻bit相加, 结果存在那2bits里(16个结果). i = (i & 0x55555555) + ((i >> 1) & 0x55555555);
再将每2bits与相邻2bits相加结果存入那4bits中.(8个结果). i = (i & 0x33333333) + ((i >> 2) & 0x33333333);
重复直到剩下一个结果为止.

代码

c++

class Solution {
public:
    int hammingWeight(uint32_t i) {
      i = (i  & 0x55555555) + ((i >> 1) & 0x55555555);
      i = (i & 0x33333333) + ((i >> 2) & 0x33333333);
      i = (i + (i >> 4)) & 0x0f0f0f0f;
      i = (i + (i >> 8)) & 0x00ff00ff;
      i = i + (i >>16) & 0x0000ffff;
      return (int)i;
    }
};