README_EN.md

2000. Reverse Prefix of Word

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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.

Solutions

Python3

class Solution:
    def reversePrefix(self, word: str, ch: str) -> str:
        i = word.find(ch)
        return word if i == -1 else word[i::-1] + word[i + 1:]

Java

class Solution {

    public String reversePrefix(String word, char ch) {
        int i = word.indexOf(ch);
        return i == -1
            ? word
            : new StringBuilder(word.substring(0, i + 1))
                .reverse()
                .append(word.substring(i + 1))
                .toString();
    }
}

C++

class Solution {
public:
    string reversePrefix(string word, char ch) {
        int i = word.find(ch);
        if (i != string::npos) reverse(word.begin(), word.begin() + i + 1);
        return word;
    }
};

Go

func reversePrefix(word string, ch byte) string {
	j := strings.IndexByte(word, ch)
	if j < 0 {
		return word
	}
	s := []byte(word)
	for i := 0; i < j; i++ {
		s[i], s[j] = s[j], s[i]
		j--
	}
	return string(s)
}

TypeScript

function reversePrefix(word: string, ch: string): string {
    let idx = word.indexOf(ch) + 1;
    if (!idx) return word;
    return [...word.substring(0, idx)].reverse().join('') + word.substring(idx);
}

Rust

impl Solution {
    pub fn reverse_prefix(word: String, ch: char) -> String {
        match word.find(ch) {
            Some(i) => word[..=i].chars().rev().collect::<String>() + &word[i + 1..],
            None => word,
        }
    }
}

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