LeetCode: Number Of 1 Bits Solution

Approach

Shift bit to the left and check on each shift

Implementation 1

As n is still passed as decimal representation, so keep dividing by 2

1var hammingWeight = function (n) {
2  let res = 0
3  while (n) {
4    res += n % 2 === 0 ? 0 : 1
5    n = Math.floor(n / 2)
6  }
7  return res
8}

Implementation 2: Pure bit manipulation

n >>>= 1 is equivalent to n = Math.floor(n / 2)

n & 1 does the AND operator on every bit, so if n is divisible by 2, result would be 0, else 1

1n:     1011
21:     0001
3n & 1: 0001 -> 1
4
5n:     1010
61:     0001
7n & 1: 0000 -> 0

1var hammingWeight = function (n) {
2  for (var res = 0; n !== 0; res += n & 1, n >>>= 1) {}
3  return res
4}