Reverse bits of a given 32 bits unsigned integer.

Example:

Input: 43261596

Output: 964176192

Explanation: 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?

给定32位无符号整数的反转位。

例：

输入：43261596 输出：964176192 说明：43261596以二进制表示为00000010100101000001111010011100，               以二进制形式返回964176192，表示为00111001011110000010100101000000。

跟进： 如果多次调用此函数，您将如何优化它？

思路： 十进制转二进制 然后存入列表，然后采用直接反转实现reserve（）

十进制转二进制

nb = bin(n)

二进制转十进制

int('10100111110',2)

逆序遍历

for i in range(len(l_nb)-1, -1, -1)

网上思路： https://blog.csdn.net/coder_orz/article/details/51705094

class Solution:
    # @param n, an integer
    # @return an integer
    def reverseBits(self, n):
        nb = bin(n)[:1:-1]
        print(nb)
        return int(nb + '0'*(32-len(nb)), 2)

if __name__ == "__main__":
    print(Solution().reverseBits(6))