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# Reverse Bits

## [Reverse Bits](https://leetcode.com/problems/reverse-bits)

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](https://en.wikipedia.org/wiki/Two%27s_complement). Therefore, in **Example 2** above, the input represents the signed integer `-3` and the output represents the signed integer `-1073741825`.

**Follow up**:

If this function is called many times, how would you optimize it?

**Example 1:**

```

Input: n = 00000010100101000001111010011100
Output:    964176192 (00111001011110000010100101000000)
Explanation: The input binary string 00000010100101000001111010011100 represents the unsigned integer 43261596, so return 964176192 which its binary representation is 00111001011110000010100101000000.
```

**Example 2:**

```

Input: n = 11111111111111111111111111111101
Output:   3221225471 (10111111111111111111111111111111)
Explanation: The input binary string 11111111111111111111111111111101 represents the unsigned integer 4294967293, so return 3221225471 which its binary representation is 10111111111111111111111111111111.
```

**Constraints:**

* The input must be a **binary string** of length `32`

## Solutions

### 🧠 Cpp

```cpp
class Solution
{
    void swap_bits(uint32_t &n, int i, int k)
    {
        bool i_value = n & (1 << i),
             k_value = n & (1 << k);

        n = (n & ~(1<<k)) | (i_value << k);
        n = (n & ~(1<<i)) | (k_value << i);
    }
public:
    uint32_t reverseBits(uint32_t n)
    {
        for(int i = 0; i< 16 ; ++i)
            swap_bits(n, i, 31-i);
        return n;
    }
};
```