Question

2000. Reverse Prefix of Word

中文文档

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

Solution 1: Simulation

First, we find the index $i$ where the character $ch$ first appears. Then, we reverse the characters from index $0$ to index $i$ (including index $i$). Finally, we concatenate the reversed string with the string starting from index $i + 1$.

The time complexity is $O(n)$, and the space complexity is $O(n)$. Here, $n$ is the length of the string $word$.

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 :]

class Solution {
  public String reversePrefix(String word, char ch) {
    int j = word.indexOf(ch);
    if (j == -1) {
      return word;
    }
    char[] cs = word.toCharArray();
    for (int i = 0; i < j; ++i, --j) {
      char t = cs[i];
      cs[i] = cs[j];
      cs[j] = t;
    }
    return String.valueOf(cs);
  }
}

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;
  }
};

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)
}

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

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,
    }
  }
}

class Solution {
  /**
   * @param String $word
   * @param String $ch
   * @return String
   */
  function reversePrefix($word, $ch) {
    $len = strlen($word);
    $rs = '';
    for ($i = 0; $i < $len; $i++) {
      $rs = $rs . $word[$i];
      if ($word[$i] == $ch) {
        break;
      }
    }
    if (strlen($rs) == $len && $rs[$len - 1] != $ch) {
      return $word;
    }
    $rs = strrev($rs);
    $rs = $rs . substr($word, strlen($rs));
    return $rs;
  }
}

Solution 2

class Solution {
  public String reversePrefix(String word, char ch) {
    int j = word.indexOf(ch);
    if (j == -1) {
      return word;
    }
    return new StringBuilder(word.substring(0, j + 1))
      .reverse()
      .append(word.substring(j + 1))
      .toString();
  }
}