190_Reverse_Bits

Leetcode 190: Reverse Bits
Detailed Video Explanation: https://youtu.be/ZW7st_pN05w

C++:
====
Method 1:
---------
uint32_t reverseBits(uint32_t n) {
        uint32_t result = 0;
        for(int i = 0; i < 32; ++i){
            result <<= 1;
            if(n&1) result++;
            n >>= 1;
        }
        return result;
    }
    
Method 2:
--------
uint32_t reverseBits(uint32_t n) {
        n = ((n & 0xffff0000) >> 16) | ((n & 0x0000ffff) << 16);
        n = ((n & 0xff00ff00) >> 8) | ((n & 0x00ff00ff) << 8);
        n = ((n & 0xf0f0f0f0) >> 4) | ((n & 0x0f0f0f0f) << 4);
        n = ((n & 0xcccccccc) >> 2) | ((n & 0x33333333) << 2);
        n = ((n & 0xaaaaaaaa) >> 1) | ((n & 0x55555555) << 1);
        return n;
    }
    
    
Java:
=====
Method 1:
---------
public int reverseBits(int n) {
        int result = 0;
        for(int i = 0; i < 32; ++i){
            result <<= 1;
            if ((n&1) > 0) result++;
            n >>= 1;
        }
        return result;
        
Method 2:
--------
public int reverseBits(int n) {
    return Integer.reverse(n);
}


Python3:
========
Method 1:
--------
def reverseBits(self, n: int) -> int:
        result = 0
        for i in range(32):
            result <<= 1
            if n&1: result += 1
            n >>= 1
        return result
        
Method 2:
--------
def reverseBits(self, n: int) -> int:
        n = ((n & 0xffff0000) >> 16) | ((n & 0x0000ffff) << 16)
        n = ((n & 0xff00ff00) >> 8) | ((n & 0x00ff00ff) << 8)
        n = ((n & 0xf0f0f0f0) >> 4) | ((n & 0x0f0f0f0f) << 4)
        n = ((n & 0xcccccccc) >> 2) | ((n & 0x33333333) << 2)
        n = ((n & 0xaaaaaaaa) >> 1) | ((n & 0x55555555) << 1)
        return n