[Leetcode/파이썬] 190. Reverse Bits

Reverse Bits

---

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?

 

class Solution:
    def reverseBits(self, n: int) -> int:
        res = 0

        for _ in range(32) :
            res = res << 1
            res += n & 1
            n >>= 1

        return res
        
        
'''디버깅용
class Solution:
    def reverseBits(self, n: int) -> int:
        res = 0

        for i in range(32) :
            res = res << 1
            res += n & 1
            print(f"[{i+1:2}] n bit: {n & 1} | res : {format(res, '032b')} | n: {format(n, '032b')}")
            n >>= 1

        return res
'''

bit 연산 방식을 채택한 풀이를 보아서 따라해보았다.

bit 연산을 알아보자. 
n<<k 왼쪽 시프트는 비트를 왼쪽으로 k 칸 옮기는 연산이다. n * (2^k)
n>>k 오른쪽 시프트는 비트를 오른쪽으로 k 칸 옮기는 연산이다. n // (2^k)
n&k 는 n과 k가 이진법으로 표기했을 때 둘 다 1이 되는 부분은 1로, 나머지는 0으로 반환하는 연산이다. 13&1 = 1101&0001 = 0001 이 된다.