[NeetCode 150] Reverse Bits

ShadyPi2024-10-22 10:07

Reverse Bits

Given a 32-bit unsigned integer n, reverse the bits of the binary representation of n and return the result.

Example 1:

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Input: n = 00000000000000000000000000010101

Output:    2818572288 (10101000000000000000000000000000)

Explanation: Reversing 00000000000000000000000000010101, which represents the unsigned integer 21, gives us 10101000000000000000000000000000 which represents the unsigned integer 2818572288.

Solution

We can convert the number into a binary number and store its bits, then sum up the bits in reverse order.

Or actually we don't need to store the bits, as we've already known where the bit will be after reversion.

Code

Convert and store:

py 复制代码

class Solution:
    def reverseBits(self, n: int) -> int:
        bits = []
        while n > 0:
            bits.append(n%2)
            n //= 2
        
        while len(bits) < 32:
            bits.append(0)

        ans = 0
        for bit in bits:
            ans <<= 1
            ans += bit
        
        return ans

No store:

py 复制代码

class Solution:
    def reverseBits(self, n: int) -> int:
        ans = 0
        for i in range(32):
            bit = (n >> i) & 1
            ans += bit<<(31-i)
        return ans