2000. Reverse Prefix of Word

Given a 0-indexed string `word` and a character `ch`, reverse the segment of `word` that starts at index `0` and ends at the index of the first occurrence of `ch` (inclusive). If the character `ch` does not exist in `word`, do nothing.

• For example, if `word = "abcdefd"` and `ch = "d"`, then you should reverse the segment that starts at `0` and ends at `3` (inclusive). The resulting string will be `"dcbaefd"`.

Return the resulting string.

Example 1:

```Input: word = "abcdefd", ch = "d"
Output: "dcbaefd"
Explanation: The first occurrence of "d" is at index 3.
Reverse the part of word from 0 to 3 (inclusive), the resulting string is "dcbaefd".
```

Example 2:

```Input: word = "xyxzxe", ch = "z"
Output: "zxyxxe"
Explanation: The first and only occurrence of "z" is at index 3.
Reverse the part of word from 0 to 3 (inclusive), the resulting string is "zxyxxe".
```

Example 3:

```Input: word = "abcd", ch = "z"
Output: "abcd"
Explanation: "z" does not exist in word.
You should not do any reverse operation, the resulting string is "abcd".
```

Constraints:

• `1 <= word.length <= 250`
• `word` consists of lowercase English letters.
• `ch` is a lowercase English letter.

2000. Reverse Prefix of Word
``````struct Solution;

impl Solution {
fn reverse_prefix(word: String, ch: char) -> String {
let n = word.len();
let a: Vec<char> = word.chars().collect();
match word.find(ch) {
Some(j) => {
let mut res = "".to_string();
for i in (0..=j).rev() {
res.push(a[i]);
}
for i in j + 1..n {
res.push(a[i]);
}
res
}
None => word,
}
}
}

#[test]
fn test() {
let word = "abcdefd".to_string();
let ch = 'd';
let res = "dcbaefd".to_string();
assert_eq!(Solution::reverse_prefix(word, ch), res);
let word = "xyxzxe".to_string();
let ch = 'z';
let res = "zxyxxe".to_string();
assert_eq!(Solution::reverse_prefix(word, ch), res);
let word = "abcd".to_string();
let ch = 'z';
let res = "abcd".to_string();
assert_eq!(Solution::reverse_prefix(word, ch), res);
}
``````