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# Reverse Bits LeetCode Solution

## Problem – 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`

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

### Reverse Bits LeetCode Solution in Java

``````public int reverseBits(int n) {
if (n == 0) return 0;

int result = 0;
for (int i = 0; i < 32; i++) {
result <<= 1;
if ((n & 1) == 1) result++;
n >>= 1;
}
return result;
}
``````

### Reverse Bits LeetCode Solution in C++

``````class Solution {
public:
uint32_t reverseBits(uint32_t n) {
n = (n >> 16) | (n << 16);
n = ((n & 0xff00ff00) >> 8) | ((n & 0x00ff00ff) << 8);
n = ((n & 0xf0f0f0f0) >> 4) | ((n & 0x0f0f0f0f) << 4);
n = ((n & 0xcccccccc) >> 2) | ((n & 0x33333333) << 2);
n = ((n & 0xaaaaaaaa) >> 1) | ((n & 0x55555555) << 1);
return n;
}
};
``````

### Reverse Bits LeetCode Solution in Python

``````def reverseBits(self, n):
ans = 0
for i in xrange(32):
ans = (ans << 1) + (n & 1)
n >>= 1
return ans``````
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