345. Reverse Vowels of a String

題目

Write a function that takes a string as input and reverse only the vowels of a string.

Example 1: Given s = "hello", return "holle".

Example 2: Given s = "leetcode", return "leotcede".

Note: The vowels does not include the letter "y".

翻譯

寫一個函式將字串的母音做反轉。

範例 1: Given s = "hello", return "holle".

範例 2: Given s = "leetcode", return "leotcede".

注意： 母音不包含 'y'。

解法

Java

Solution 1.

這題將母音位置反轉其餘不變，所以用一個會圈分別從頭和尾巴同時一個一個掃，找到母音即可交換位置，這裡要注意的是字串字母包含大小寫所以母音也要同時判斷他的大小寫。

• String 走訪
• Run Time: 17 ms
• 時間複雜度: O(n log n)
• 空間複雜度: O(n)

class Solution {
    public String reverseVowels(String s) {
        String vowels = "aeiouAEIOU";
        char arr[] = s.toCharArray();
        int i = 0, j = s.length() - 1;
        while (i < j) {
            while (i < j && !vowels.contains(arr[i] + "")) {
                i++;
            }
            while (i < j && !vowels.contains(arr[j] + "")) {
                j--;
            }
            char temp = arr[i];
            arr[i] = arr[j];
            arr[j] = temp;
            i++;
            j--;
        }
        return new String(arr);
    }
}

C

Solution 1.

C 比較麻煩指標陣列不能直接替換值，所以要先複製到一般的字元陣列中，再逐一的從頭和尾巴同時尋找母音字母。

• String 走訪
• Run Time: 6 ms
• 時間複雜度: O(n log n)
• 空間複雜度: O(n)

char * reverseVowels(char * s) {
  int i = 0, j = strlen(s) - 1;
  char arr[1000000];
  strcpy(arr, s);
  char * ans = arr;
  while (i < j) {
    while (i < j && !(s[i] == 'a' || s[i] == 'A' || s[i] == 'e' || s[i] == 'E' || s[i] == 'i' || s[i] == 'I' || s[i] == 'o' || s[i] == 'O' || s[i] == 'u' || s[i] == 'U')) {
      arr[i] = s[i];
      i++;
    }
    while (i < j && !(s[j] == 'a' || s[j] == 'A' || s[j] == 'e' || s[j] == 'E' || s[j] == 'i' || s[j] == 'I' || s[j] == 'o' || s[j] == 'O' || s[j] == 'u' || s[j] == 'U')) {
      arr[j] = s[j];
      j--;
    }
    arr[j] = s[i];
    arr[i] = s[j];
    i++;
    j--;
  }
  return ans;
}