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?

Related problem: Reverse Integer

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

这道题又是在考察位操作Bit Operation，LeetCode中有关位操作的题也有不少，比如 Repeated DNA Sequences，Single Number,   Single Number II ，和 Grey Code 等等。跟上面那些题比起来，这道题简直不能再简单了。那么对于这道题，我们只需要把要翻转的数从右向左一位位的取出来，如果取出来的是1，我们将结果res左移一位并且加上1；如果取出来的是0，我们将结果res左移一位，然后将n右移一位即可，代码如下：

解法一：

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

我们可以简化上面的代码，去掉if…else…结构，可以结果res左移一位，然后再判断n的最低位是否为1，是的话那么结果res加上1，然后将n右移一位即可，代码如下：

解法二：

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

我们继续简化上面的解法，将if判断句直接揉进去，通过‘或’上一个n的最低位即可，用n‘与’1提取最低位，然后将n右移一位即可，代码如下：

解法三：

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

博主还能进一步简化，这里我们不更新n的值，而是直接将n右移i位，然后通过‘与’1来提取出该位，加到左移一位后的结果res中即可，参加代码如下：

解法四：

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

我们也可以换一种角度来做，首先将n右移i位，然后通过‘与’1来提取出该位，然后将其左移 (32 – i) 位，然后‘或’上结果res，就是其翻转后应该在的位置，参见代码如下： 

解法五：

class Solution {
public:
    uint32_t reverseBits(uint32_t n) {
        uint32_t res = 0;
        for (int i = 0; i < 32; ++i) {
            res |= ((n >> i) & 1) << (31 - i);
        }
        return res;
    }
};

讨论：这道题的最高票解法实在是很叼啊，参见这个帖子，但是博主没有太理解啊，希望哪位大神能讲解一下哈～

类似题目：

Number of 1 Bits

参考资料：

https://leetcode.com/problems/reverse-bits/discuss/54938/A-short-simple-Java-solution

https://leetcode.com/problems/reverse-bits/discuss/54772/The-concise-C++-solution(9ms)

https://leetcode.com/problems/reverse-bits/discuss/54741/O(1)-bit-operation-C++-solution-(8ms)

https://leetcode.com/problems/reverse-bits/discuss/54738/Sharing-my-2ms-Java-Solution-with-Explanation

https://leetcode.com/problems/reverse-bits/discuss/54873/Java-two-methods-using-String-or-bit-operation-6ms-and-2ms-easy-understand