Reverse the Bits of a 32-bit Integer

easy bit manipulation

Problem

Given a 32-bit unsigned integer n, return its bitwise reverse — bit i of the input becomes bit 31 − i of the output.

Inputn = 0b00000010100101000001111010011100

Output0b00111001011110000010100101000000

Loop 32 times. Each step shifts the answer left, ORs in the lowest bit of n, then shifts n right.

n (decimal, 0 ≤ n < 2^32)

def reverse_bits(n):
    result = 0
    for _ in range(32):
        result = (result << 1) | (n & 1)
        n >>= 1
    return result

function reverseBits(n) {
  let result = 0;
  for (let i = 0; i < 32; i++) {
    result = (result << 1) | (n & 1);
    n = n >>> 1;
  }
  return result >>> 0;
}

class Solution {
    public int reverseBits(int n) {
        int result = 0;
        for (int i = 0; i < 32; i++) {
            result = (result << 1) | (n & 1);
            n >>>= 1;
        }
        return result;
    }
}

uint32_t reverseBits(uint32_t n) {
    uint32_t result = 0;
    for (int i = 0; i < 32; i++) {
        result = (result << 1) | (n & 1u);
        n >>= 1;
    }
    return result;
}

Time: O(32) = O(1) Space: O(1)