# Question

Formatted question description: https://leetcode.ca/all/191.html

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?

# Algorithm

Sum up &1 result.

# Code

• 
public class Number_of_1_Bits {

public static void main(String[] args) {
Number_of_1_Bits out = new Number_of_1_Bits();

Solution s = out.new Solution();

System.out.println(s.hammingWeight(11));
}

public class Solution {
// you need to treat n as an unsigned value
public int hammingWeight(int n) {

int count = 0;

while (n > 0) {
count += n & 1;

n >>= 1;
}
return count;
}
}

public class Solution_cheat {
// you need to treat n as an unsigned value
public int hammingWeight(int n) {
return Integer.bitCount(n); // Returns the number of one-bits in the two's complement binary representation
}
}

public class Solution_toBinaryString {
// you need to treat n as an unsigned value
public int hammingWeight(int n) {

String binaryString = Integer.toBinaryString(n);

int count = 0;
for (int i = 0; i < binaryString.length(); i++) {
if (binaryString.charAt(i) == '1') {
count++;
}
}

return count;
}
}
}

############

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;
}
}

• // OJ: https://leetcode.com/problems/number-of-1-bits/
// Time: O(1)
// Space: O(1)
class Solution {
public:
int hammingWeight(uint32_t n) {
int ans = 0;
for (; n; n -= (n & -n)) ++ans;
return ans;
}
};

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

############

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


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