carloscn
commented
1 year ago Analysis
pub fn reverse_prefix(word: &str, ch: char) -> String
{
if word.len() < 1 {
return String::new();
}
let mut ret:Vec<char> = vec![];
let mut t_v:Vec<char> = vec![];
let w_v:Vec<char> = word.chars().collect();
for i in 0..word.len() {
if w_v[i] != ch {
t_v.push(w_v[i]);
} else {
t_v.push(w_v[i]);
break;
}
}
if t_v.len() == w_v.len() {
return w_v.iter().collect();
}
for i in 0..word.len() {
if !t_v.is_empty() {
ret.push(t_v.pop().unwrap());
continue;
}
ret.push(w_v[i])
}
return ret.iter().collect();
}
Description
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.