190. Reverse Bits

最新推荐文章于 2024-01-19 11:13:02 发布

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最新推荐文章于 2024-01-19 11:13:02 发布

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本文介绍了一种算法，用于反转一个给定的32位无符号整数的二进制位，并提供了两种实现方式。第一种方法通过字符串操作来完成反转，第二种方法则利用位操作进行优化。

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190. Reverse Bits

Leetcode link for this question

Discription:

Reverse bits of a given 32 bits unsigned integer.

For example, given input 43261596 (represented in binary as 00000010100101000001111010011100), return 964176192 (represented in binary as 00111001011110000010100101000000).

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

Analyze:

Code 1:

class Solution(object):
    def reverseBits(self, n):
        """
        :type n: int
        :rtype: int
        """
        bi=str(bin(n))[2:]
        bi='0'*(32-len(bi))+bi      
        s=0 
        wei=1
        for i in bi:                                 
            s=int(i)*wei+s
            wei*=2
        return s

Submission Result:

Status: Accepted
Runtime: 68 ms
Ranking: beats 31.14%

Code 2:

class Solution(object):
    def reverseBits(self, n):
        """
        :type n: int
        :rtype: int
        """
        for i in range(16):
            j=32-i-1
            if i>j:
                break
            bi=(n>>i)&1
            bj=(n>>j)&1
            if bi^bj:          
                n=n^(1<<i)^(1<<j)
        return n