Reverse Bits

Description

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?

Code

Time Complexity: $O(n)$, Space Complexity: $O(n)$

運算優先級：>> 優先於 &。

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

class Solution {
public:
    uint32_t reverseBits(uint32_t n) {
        uint32_t reversed = 0;
        for(unsigned int i = 0, mask = 1; i < 32; i++, mask <<= 1) {
            if(n & mask)
                reversed |= (1 << (32 - 1 - i)); 
        }
 
        return reversed;
    }
};

class Solution {
public:
    uint32_t reverseBits(uint32_t n) {
        uint32_t answer = 0;
        for(int i = 0; i < 32; i++) {
            answer = answer << 1;
            answer = answer ^ (n & 1);
            n = n >> 1;
        }
 
        return answer;
    }
};

Time Complexity: $O(\log n)$, Space Complexity: $O(n)$

運用到線性代數的觀念：$(AB{)}^T=B^T{A}^T$

以 4 位元來當例子：abcd → cdab → dcba。底下的 code 就是針對每一個 bit 去做同樣的事情。

class Solution {
public:
    uint32_t reverseBits(uint32_t n) {
        n = (n >> 16) | (n << 16);
        n = ((n & 0xff00ff00) >> 8) | ((n & 0x00ff00ff) << 8);
        n = ((n & 0xf0f0f0f0) >> 4) | ((n & 0x0f0f0f0f) << 4);
        n = ((n & 0xcccccccc) >> 2) | ((n & 0x33333333) << 2);
        n = ((n & 0xaaaaaaaa) >> 1) | ((n & 0x55555555) << 1);
        return n;
    }
};

Source

• Reverse Bits - LeetCode