Given a positive integer `n`

, find and return *the longest distance between any two adjacent *

`1`

`n`

`1`

`0`

Two `1`

's are **adjacent** if there are only `0`

's separating them (possibly no `0`

's). The **distance** between two `1`

's is the absolute difference between their bit positions. For example, the two `1`

's in `"1001"`

have a distance of 3.

**Example 1:**

Input:n = 22Output:2Explanation:22 in binary is "10110". The first adjacent pair of 1's is "10110" with a distance of 2. The second adjacent pair of 1's is "10110" with a distance of 1. The answer is the largest of these two distances, which is 2. Note that "10110" is not a valid pair since there is a 1 separating the two 1's underlined.

**Example 2:**

Input:n = 5Output:2Explanation:5 in binary is "101".

**Example 3:**

Input:n = 6Output:1Explanation:6 in binary is "110".

**Example 4:**

Input:n = 8Output:0Explanation:8 in binary is "1000". There aren't any adjacent pairs of 1's in the binary representation of 8, so we return 0.

**Example 5:**

Input:n = 1Output:0

**Constraints:**

`1 <= n <= 10`

^{9}

```
struct Solution;
impl Solution {
fn binary_gap(n: i32) -> i32 {
let mut max = 0;
let mut prev: Option<usize> = None;
for i in 0..32 {
let bit = 1 << i;
if n & bit != 0 {
if let Some(j) = prev {
max = usize::max(i - j, max);
}
prev = Some(i);
}
}
max as i32
}
}
#[test]
fn test() {
assert_eq!(Solution::binary_gap(22), 2);
assert_eq!(Solution::binary_gap(5), 2);
assert_eq!(Solution::binary_gap(6), 1);
assert_eq!(Solution::binary_gap(8), 0);
}
```