# 191. Number of 1 Bits

## Description

Write a function that takes the binary representation of 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, the input represents the signed integer. -3.

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.

Follow up: If this function is called many times, how would you optimize it?

## Solutions

• public class Solution {
// you need to treat n as an unsigned value
public int hammingWeight(int n) {
int ans = 0;
while (n != 0) {
n &= n - 1;
++ans;
}
return ans;
}
}

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

• class Solution:
def hammingWeight(self, n: int) -> int:
ans = 0
while n:
n &= n - 1
ans += 1
return ans


• func hammingWeight(num uint32) int {
ans := 0
for num != 0 {
num &= num - 1
ans++
}
return ans
}

• function hammingWeight(n: number): number {
let ans: number = 0;
while (n !== 0) {
ans++;
n &= n - 1;
}
return ans;
}


• /**
* @param {number} n - a positive integer
* @return {number}
*/
var hammingWeight = function (n) {
let ans = 0;
while (n) {
n &= n - 1;
++ans;
}
return ans;
};


• impl Solution {
pub fn hammingWeight(n: u32) -> i32 {
n.count_ones() as i32
}
}