190. Reverse Bits

Problem

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. 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?

Solution

Bit Manipulation - O(1) time - O(1) space

The reason this is O(1) time instead of O(n) is we iterate a fixed number of times. It’s really O(32) which reduces to O(1)

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

We can also replace the Bitwise OR with an addition to the same effect

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