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Top Interview Questions 의 Easy Collection에 있는 문제입니다.
문제는 여기서 볼 수 있습니다.
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?
- 비트 전체를 순회하면서 1 과 AND 연산을 반복합니다.
- n 의 가장 오른쪽에 있는 값이 answer 의 가장 왼쪽에 가야 하므로 answer 는 0 부터 왼쪽 쉬프트, n 은 오른쪽 쉬프트를 반복하며 계속 비교하면 됩니다.
class Solution:
def reverseBits(self, n: int) -> int:
answer = 0
# Constraints: The input must be a binary string of length 32
for i in range(32):
answer = (answer << 1) + (n & 1)
n >>= 1
return answer
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