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[技术交流] 【朱迪的LeetCode刷题笔记】】190. Reverse Bits #Easy

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发表于 2023-6-7 10:09:04 | 显示全部楼层 |阅读模式

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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?


Sol1
https://leetcode.com/problems/re ... in-built-functions/
  1. class Solution:
  2.     def reverseBits(self, n: int) -> int:
  3.         res=0
  4.         for i in range(32): # since it is 32 bit integer
  5.             res=res<<1   # left shift to check it till 32 bits
  6.             res+=n%2  # checking if rightmost bit of n is 1 and adding to result
  7.             n=n>>1     # checking all bits of n till n becomes 0
  8.         return res
复制代码


Sol2
https://leetcode.com/problems/re ... c-with-63ms-3lines/

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