190. Reverse Bits

Difficulty:
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Problem

Reverse bits of a given 32 bits unsigned integer.

Note:

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:

Solution (Java)

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;
    }
}

Solution (Javascript)

/**
 * @param {number} n - a positive integer
 * @return {number} - a positive integer
 */
var reverseBits = function(n) {
    //convert the number to base2 then stringify, split, and reverse 
    let reversedArray = n.toString(2).split("").reverse()
    //the converted number may not be 32 digits so pad the end 
    while(reversedArray.length <32){ reversedArray.push('0')}
    //join the string, then convert to Integer from base 2
    return  parseInt(reversedArray.join(""),2) 
};

Explain:

nope.