Given a string `s`

and a character `c`

that occurs in `s`

, return *an array of integers *`answer`

* where *`answer.length == s.length`

* and *`answer[i]`

* is the distance from index *

`i`

`c`

`s`

.The **distance** between two indices `i`

and `j`

is `abs(i - j)`

, where `abs`

is the absolute value function.

**Example 1:**

Input:s = "loveleetcode", c = "e"Output:[3,2,1,0,1,0,0,1,2,2,1,0]Explanation:The character 'e' appears at indices 3, 5, 6, and 11 (0-indexed). The closest occurrence of 'e' for index 0 is at index 3, so the distance is abs(0 - 3) = 3. The closest occurrence of 'e' for index 1 is at index 3, so the distance is abs(1 - 3) = 3. For index 4, there is a tie between the 'e' at index 3 and the 'e' at index 5, but the distance is still the same: abs(4 - 3) == abs(4 - 5) = 1. The closest occurrence of 'e' for index 8 is at index 6, so the distance is abs(8 - 6) = 2.

**Example 2:**

Input:s = "aaab", c = "b"Output:[3,2,1,0]

**Constraints:**

`1 <= s.length <= 10`

^{4}`s[i]`

and`c`

are lowercase English letters.- It is guaranteed that
`c`

occurs at least once in`s`

.

```
struct Solution;
impl Solution {
fn shortest_to_char(s: String, c: char) -> Vec<i32> {
let s: Vec<char> = s.chars().collect();
let mut prev: Option<usize> = None;
let n = s.len();
let mut res: Vec<i32> = vec![n as i32; n];
for i in 0..n {
if s[i] == c {
prev = Some(i);
}
if let Some(j) = prev {
res[i] = i32::min(res[i], (i - j) as i32);
}
}
prev = None;
for i in (0..n).rev() {
if s[i] == c {
prev = Some(i);
}
if let Some(j) = prev {
res[i] = i32::min(res[i], (j - i) as i32);
}
}
res
}
}
#[test]
fn test() {
let s = "loveleetcode".to_string();
let t = vec![3, 2, 1, 0, 1, 0, 0, 1, 2, 2, 1, 0];
assert_eq!(Solution::shortest_to_char(s, 'e'), t);
}
```