5/19 每日一題 (反轉32位無符號整數)

 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:

        result=0

        for _ in range(32): #處理32次,32位無符號整數

            result= (result<<1) | (n&1)

            n >>= 1

    #運算符順序:先處理左位元移動(<<) 在來是位元與(&) 最後位元或(|)

    #但是有括號所以優先處理括號內

    #(res<<1) 和 (n&1) 會先處理完成後再進行|運算

    #因為res是0 ,所以res執行完res<<1後可以確定 res|n&1 必定是 0|&1

        return result -------------------------------------

參考解答

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