LeetCode-in-Java

190. Reverse Bits

Easy

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:

Follow up: If this function is called many times, how would you optimize it?

Solution

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;
    }
}