LeetCode Reverse Bits

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?
Related problem: Reverse Integer

居然没有记录，不过买了本数全是位操作，真好可以看一下，先来个常规解法

class Solution {
public:
    uint32_t reverseBits(uint32_t n) {
        uint32_t res = 0;
        for (int i=0; i<32; i++) {
            res = (res << 1) | (n & 0x1);
            n>>=1;
        }
        return res;
    }
};

古典解法

class Solution {
public:
    uint32_t reverseBits(uint32_t n) {
        n = (n & 0x55555555) << 1 | (n & 0xAAAAAAAA) >> 1;
        n = (n & 0x33333333) << 2 | (n & 0xCCCCCCCC) >> 2;
        n = (n & 0x0F0F0F0F) << 4 | (n & 0xF0F0F0F0) >> 4;
        n = (n & 0x00FF00FF) << 8 | (n & 0xFF00FF00) >> 8;
        n = (n & 0x0000FFFF) << 16| (n & 0xFFFF0000) >> 16;
        return n;
    }
};

以上代码来自《算法心得》