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Java-based LeetCode algorithm problem solutions, regularly updated

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package g0101_0200.s0190_reverse_bits;

// #Easy #Top_Interview_Questions #Bit_Manipulation #Divide_and_Conquer
// #Algorithm_I_Day_14_Bit_Manipulation #Udemy_Bit_Manipulation
// #2022_06_27_Time_1_ms_(98.66%)_Space_41.9_MB_(81.78%)

/**
 * 190 - Reverse Bits\.
 *
 * Easy
 *
 * Reverse bits of a given 32 bits unsigned integer.
 *
 * **Note:**
 *
 * *   Note that in some languages, such as Java, there is no unsigned integer type. In this case, both input and output will be given as a signed integer type. They should not affect your implementation, as the integer's internal binary representation is the same, whether it is signed or unsigned.
 * *   In Java, the compiler represents the signed integers using [2's complement notation](https://en.wikipedia.org/wiki/Two%27s_complement). Therefore, in **Example 2** above, the input represents the signed integer `-3` and the output represents the signed integer `-1073741825`.
 *
 * **Example 1:**
 *
 * **Input:** n = 00000010100101000001111010011100
 *
 * **Output:** 964176192 (00111001011110000010100101000000)
 *
 * **Explanation:** The input binary string **00000010100101000001111010011100** represents the unsigned integer 43261596, so return 964176192 which its binary representation is **00111001011110000010100101000000**. 
 *
 * **Example 2:**
 *
 * **Input:** n = 11111111111111111111111111111101
 *
 * **Output:** 3221225471 (10111111111111111111111111111111)
 *
 * **Explanation:** The input binary string **11111111111111111111111111111101** represents the unsigned integer 4294967293, so return 3221225471 which its binary representation is **10111111111111111111111111111111**. 
 *
 * **Constraints:**
 *
 * *   The input must be a **binary string** of length `32`
 *
 * **Follow up:** If this function is called many times, how would you optimize it?
**/
public class Solution {
    // you need treat n as an unsigned value
    public int reverseBits(int n) {
        int ret = 0;
        // because there are 32 bits in total
        for (int i = 0; i < 32; i++) {
            ret = ret << 1;
            // If the bit is 1 we OR it with 1, ie add 1
            if ((n & 1) > 0) {
                ret = ret | 1;
            }
            n = n >>> 1;
        }
        return ret;
    }
}