博文中会简要介绍Leetcode P0191题目分析及解题思路。

“Number of 1 Bits”是一道比较简单的涉及位操作的题目。这道题其实基本思路和p0190一致，都是向右移动比特位，每次查看末尾是0还是1，根据末尾比特位上的值来累加。

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

Example 1:

Input: 00000000000000000000000000001011
Output: 3
Explanation: The input binary string 00000000000000000000000000001011 has a total of three '1' bits.

Example 2:

Input: 00000000000000000000000010000000
Output: 1
Explanation: The input binary string 00000000000000000000000010000000 has a total of one '1' bit.

Example 3:

Input: 11111111111111111111111111111101
Output: 31
Explanation: The input binary string 11111111111111111111111111111101 has a total of thirty one '1' bits.

Note:

• Note that in some languages such as Java, there is no unsigned integer type. In this case, the input will be given as signed integer type and should not affect your implementation, as the internal binary representation of the integer 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?

以下是Java的题解代码实现。

public class Solution {
    // you need to treat n as an unsigned value
    public int hammingWeight(int n) {
        
        int ones = 0, count = 0;
        while (n != 0 && count < 32) {
            ones += n&1;
            n >>= 1;
            ++count;
        }
        
        return ones;
    }
}

以下是C++的题解代码实现。

class Solution {
public:
    int hammingWeight(uint32_t n) {
        int ones = 0, count = 0;
        while (n != 0 && count < 32) {
            ones += n&1;
            n >>= 1;
            ++count;
        }
        
        return ones;
    }
};