###### tags: `Leetcode` `easy` `bit` `python` `c++`

# 190. Reverse Bits

## [題目來源:] https://leetcode.com/problems/reverse-bits/

## 題目:
Reverse bits of a given 32 bits unsigned integer.

Note:

* Note that in some languages, such as Java, there is no unsigned integer type. In this case, both input and output will be given as a signed integer type. They should not affect your implementation, as the integer's internal binary representation is the same, whether it is signed or unsigned.
* In Java, the compiler represents the signed integers using 2's complement notation. Therefore, in Example 2 above, the input represents the signed integer -3 and the output represents the signed integer -1073741825.

#### [圖片來源:] https://leetcode.com/problems/reverse-bits/

## 解題想法:
兄弟題目: [P191. Number of 1 Bits](/R4rPTOWjTWSS0ZfMkz6A9g)

' >> ' 和 ' << ' 都是位運算，對二進制數進行移位操作
ex: <<2 二進位補兩個0 即乘4
ex: 5 = 101 ' <<2 ' == 10100 == 20

逐一取出最後一位數添加到最左邊高位元即可

## Python:

class Solution:
    # @param n, an integer
    # @return an integer
    
    def reverseBits(self, n):
        res=0
        power=31
        while n:
            #get最右邊的值 用(&1)
            last=n&1
            res+=last<<power
            n>>=1
            power-=1
        return res
        '''
        b = bin(n)[:1:-1]
        return int(b+'0'*(32-len(b)),2)
        '''

if __name__ == '__main__':
    result = Solution()
    n = int('00000010100101000001111010011100', 2)
    ans = result.reverseBits(n)
    print(n,ans)

## C++:
uint32_t為0~2^32-1 (0x00000000~0xFFFFFFFF)

#include<iostream>
using namespace std;

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

int main(){
    Solution res;
    uint32_t n= 43261596;  //binary: 00000010100101000001111010011100
    cout<<res.reverseBits(n)<<endl;
    return 0;
}