190. Reverse Bits

/**
 * Problem: LeetCode 190 - Reverse Bits
 * Difficulty: Easy
 * Language: Java
 * Author: CodeByTushu
 */
public class Solution {
    // you need treat n as an unsigned value
    public int reverseBits(int n) {
        int result = 0;
        for (int i = 0; i < 32; i++) {
            result = (result << 1) | (n & 1);
            n >>>= 1;
        }
        return result;
    }
}

"""
Problem: LeetCode 190 - Reverse Bits
Difficulty: Easy
Language: Python
Author: CodeByTushu
"""
class Solution:
    def reverseBits(self, n: int) -> int:
        result = 0
        for _ in range(32):
            result = (result << 1) | (n & 1)
            n >>= 1
        return result

/**
 * Problem: LeetCode 190 - Reverse Bits
 * Difficulty: Easy
 * Language: C++
 * Author: CodeByTushu
 */
#include <cstdint>

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

/**
 * Problem: LeetCode 190 - Reverse Bits
 * Difficulty: Easy
 * Language: JavaScript
 * Author: CodeByTushu
 */
/**
 * @param {number} n - a positive integer
 * @return {number} - a positive integer
 */
var reverseBits = function(n) {
    let result = 0;
    for (let i = 0; i < 32; i++) {
        result = (result << 1) | (n & 1);
        n >>>= 1;
    }
    return result >>> 0;
};