# Reverse Bits LeetCode Solution | Easy Approach Share:

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. Therefore, in Example 2 above, the input represents the signed integer `-3` and the output represents the signed integer `-1073741825`.

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`

Time: O(32)=O(1)
Space: O(1)

### C++

`````` class Solution {
public:
uint32_t reverseBits(uint32_t n) {
uint32_t ans = 0;

for (int i = 0; i < 32; ++i)
if (n >> i & 1)
ans |= 1 << 31 - i;

return ans;
}
};
``````

### Java

``````
public class Solution {
// you need treat n as an unsigned value
public int reverseBits(int n) {
int ans = 0;

for (int i = 0; i < 32; ++i)
if ((n >> i & 1) == 1)
ans |= 1 << 31 - i;

return ans;
}
}
``````

### Python

``````class Solution:
def reverseBits(self, n: int) -> int:
ans = 0

for i in range(32):
if n >> i & 1:
ans |= 1 << 31 - i

return ans

``````

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