题目描述：

everse 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.

解题思路：

将一个整形数的二进制序列逆转，后输出二进制序列代表的十进制数

方法一：

采用一个32长度的数组，存储二进制，之后逆转，再计算此数组所表示的整形数。

class Solution {
public:
    uint32_t reverseBits(uint32_t n) {
        vector<int> a(32,0);
	int mood=0,i=0,temp;
	unsigned int temp1=0;
	while(n/2!=0)
	{
		mood=n%2;
		a[i]=mood;
		n/=2;
		i++;
	}
	a[i]=n%2;
	reverse(a.begin(),a.end());
	for(int j=0;j<32;j++)
	{
		if (a[j]!=0)
		{
			temp=pow(2*1.0,j);
			temp1+=temp;
		}
	}
	return temp1;
    }
};

方法二：

利用位运算符，从低位到高位取出原整形数的各个比特位，对应转换为0~31，1~30，...，31~0。用31-当前下标 就是转换之后的下标，在用左移操作即可实现二进制到十进制的转换。

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