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).

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`

191. Number of 1 Bits
``````#![allow(clippy::unreadable_literal)]
struct Solution;

impl Solution {
#[allow(non_snake_case)]
fn hammingWeight(n: u32) -> i32 {
n.count_ones() as i32
}
}

#[test]
fn test() {
let n = 0b00000000000000000000000000001011;
let res = 3;
assert_eq!(Solution::hammingWeight(n), res);
let n = 0b00000000000000000000000010000000;
let res = 1;
assert_eq!(Solution::hammingWeight(n), res);
let n = 0b11111111111111111111111111111101;
let res = 31;
assert_eq!(Solution::hammingWeight(n), res);
}
``````