Reverse Bits (leetcode 190)

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

Credits:
Special thanks to @ts for adding this problem and creating all test cases.

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用位运算，将第一位和最后一位反转。

网上牛人方法：
把一个32位整数按位反转，即第1位转到第32位，第2位转到第31位，依次下去。一牛人写的算法如下：

unsigned int bit_reverse(unsigned int n) {     n = ((n >> 1) & 0x55555555) | ((n << 1) & 0xaaaaaaaa);     n = ((n >> 2) & 0x33333333) | ((n << 2) & 0xcccccccc);     n = ((n >> 4) & 0x0f0f0f0f) | ((n << 4) & 0xf0f0f0f0);     n = ((n >> 8) & 0x00ff00ff) | ((n << 8) & 0xff00ff00);     n = ((n >> 16) & 0x0000ffff) | ((n << 16) & 0xffff0000);     return n; } 

第一行代码为奇偶位相互交换；第二行为以两位为一单元，奇偶单元进行交换；第三行为以四位为一单元，奇偶单元进行交换；第四行为以八位为一单元，奇偶单元进行交换；最后一行为以十六位为一单元，奇偶单元进行交换。至此，32位反转完成，算法结束。

class Solution {public:    uint32_t reverseBits(uint32_t n) {        for (int i = 0; i < 16; ++i) {            uint32_t r = ((n & (1 << i)) == 0 ? (~(1 << (31 - i))) & n & (~(1 << i)) : ((1 << (31 - i)) | n) & (~(1 << i)));            uint32_t l = ((n & (1 << (31 - i))) == 0 ? (~(1 << i)) & n & (~(1 << (31 - i))) : ((1 << i) | n) & (~(1 << (31 - i))));            n = r | l;            cout << n << endl;        }        return n;    }};