190. Reverse Bits

https://leetcode.com/problems/reverse-bits/

Python

class Solution:
    def reverseBits(self, n: int) -> int:
        result = 0
        for power in range(31, -1, -1):
            result += (n & 1) << power
            n = n >> 1
        return result

Javascript

var reverseBits = function(n) {
  const list = [];
  for (let i = 0; i < 32; i++) {
    if (n > 0) {
      list.push(parseInt(n % 2));
      n = parseInt(n / 2)
    } else list.push(0);
  }
  return parseInt(list.join(''), 2);
};