Skip to content

345. Reverse Vowels of a String

  • Time: $O(n)$
  • Space: $O(1)$
 1
 2
 3
 4
 5
 6
 7
 8
 9
10
11
12
13
14
15
16
17
18
19
20
21
22
23
class Solution {
 public:
  string reverseVowels(string s) {
    int l = 0;
    int r = s.length() - 1;

    while (l < r) {
      while (l < r && !isVowel(s[l]))
        ++l;
      while (l < r && !isVowel(s[r]))
        --r;
      swap(s[l++], s[r--]);
    }

    return s;
  }

 private:
  bool isVowel(char c) {
    static constexpr string_view kVowels = "aeiouAEIOU";
    return kVowels.find(c) != string_view::npos;
  }
};

 1
 2
 3
 4
 5
 6
 7
 8
 9
10
11
12
13
14
15
16
17
18
19
20
21
22
23
24
class Solution {
  public String reverseVowels(String s) {
    StringBuilder sb = new StringBuilder(s);
    int l = 0;
    int r = s.length() - 1;

    while (l < r) {
      while (l < r && !isVowel(sb.charAt(l)))
        ++l;
      while (l < r && !isVowel(sb.charAt(r)))
        --r;
      sb.setCharAt(l, s.charAt(r));
      sb.setCharAt(r, s.charAt(l));
      ++l;
      --r;
    }

    return sb.toString();
  }

  private boolean isVowel(char c) {
    return "aeiouAEIOU".indexOf(c) != -1;
  }
}

 1
 2
 3
 4
 5
 6
 7
 8
 9
10
11
12
13
14
15
16
17
class Solution:
  def reverseVowels(self, s: str) -> str:
    chars = list(s)
    kVowels = 'aeiouAEIOU'
    l = 0
    r = len(s) - 1

    while l < r:
      while l < r and chars[l] not in kVowels:
        l += 1
      while l < r and chars[r] not in kVowels:
        r -= 1
      chars[l], chars[r] = chars[r], chars[l]
      l += 1
      r -= 1

    return ''.join(chars)