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);
}