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# 0190. Reverse Bits

<https://leetcode.com/problems/reverse-bits>

## Description

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`.

**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?

## ac

careful unsigned integer

```java
public class Solution {
    // you need treat n as an unsigned value
    public int reverseBits(int n) {
        // edge case
        if (n == 0) return 0;

        int res = 0;
        for (int i = 0; i < 32; i++) {
            res <<= 1;
            if ((n & 1) == 1) res++;  // cannot use n%2, unsigned integer can be larger than Integer.MAX_VALUE
            n >>= 1;
        }
        return res;
    }
}
```