[leetcode] 190. 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 as00111001011110000010100101000000).

Follow up:

If this function is called many times, how would you optimize it?
这道题是按bit位倒序数字，题目难度为Easy。

比较直观的想法是通过移位操作来获取倒序后的数字，思路比较简单就直接上代码了。具体代码：

class Solution {
public:
uint32_t reverseBits(uint32_t n) {
uint32_t ret = 0;
for(int i=0; i<32; ++i) {
ret <<= 1;
ret |= (n%2);
n >>= 1;
}
return ret;
}
};

题目追问如果多次调用该如何优化，没想到合适的方法，所以看了下别人的代码，思路比较巧妙，通过位操作来实现上述功能，避免了循环，效率有所提高。具体想法举个栗子大家就清楚了，例如32bit数字0x12345678，通过位操作逐步变化如下：

                            0x12345678

                            0x56781234

                            0x78563412

                            0x87654321

                            0x2d951c84

                            0x1e6a2c48

具体代码：

class Solution {
public:
uint32_t reverseBits(uint32_t n) {
n = ((n & 0xffff0000) >> 16) | ((n & 0x0000ffff) << 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;
}
};