Reverse Bits

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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?

思路：由于把一个数转换为二进制数的过程是反序相除法！最先的到的余数刚好就是反向后开始的数字，于是一边把这个数转换为二进制，一般对余数求和乘二即可得到最后答案

代码：

class Solution {
public:
    uint32_t reverseBits(uint32_t n) {
        int count=0;
        uint32_t ans=0;
        while(count<31){
            ans+=(n%2);
            n/=2;
            ans*=2;
            count++;
        }
        ans+=(n%2);
        return ans;
    }
};