[LeetCode Road] Number of 1 Bits - Solution/C++

191. Number of 1 Bits

Question:

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:

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

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

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

Source code

Version 1

Idea:
The question is very simple if you have understood 190. Reverse Bits.

Before shifting right 1 bit, we only need to do n & 1 to get each bit at rightmost. Use a counter variable to count how many number of 1 bits.

class Solution {
public:
    int hammingWeight(uint32_t n) {
        int count = 0;
        for(int i = 0; i < 32; i++){
            count += (n & 1 == 1)?1:0;
            n >>= 1;
        }
        return count;
    }
};