LeetCode //C - 190. Reverse Bits

190. Reverse Bits

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

From: LeetCode
Link: 190. Reverse Bits

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Solution:

Ideas：

• Initializes a result variable reversed to 0.
• Iterates 32 times, corresponding to the 32 bits of an unsigned integer.
• In each iteration, it shifts reversed to the left to make room for the next bit.
• It then takes the least significant bit of n by performing n & 1 and ORs it with reversed.
• Then it shifts n to the right by one, to process the next bit in the next iteration.

Code:

uint32_t reverseBits(uint32_t n) {
    uint32_t reversed = 0;
    for (int i = 0; i < 32; i++) {
        reversed = (reversed << 1) | (n & 1);
        n >>= 1;
    }
    return reversed;
}
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