[LeetCode] 345. Reverse Vowels of a String

Given a string s, reverse only all the vowels in the string and return it.

The vowels are 'a', 'e', 'i', 'o', and 'u', and they can appear in both cases.

Example 1:

Input: s = "hello"
Output: "holle"

Example 2:

Input: s = "leetcode"
Output: "leotcede"

Constraints:

1 <= s.length <= 3 * 105
s consist of printable ASCII characters.

反转字符串中的元音字母。

思路基本同344题。不同之处在于344题要求不能使用额外空间，但是本题非得使用额外空间才能解决，不仅需要 hashset，也需要将 input 先转成数组才能判断。同时注意这道题的 testcase 是同时包含大写字母和小写字母的。

时间O(n)

空间O(n) - hashset

JavaScript实现

 1 /**
 2  * @param {string} s
 3  * @return {string}
 4  */
 5 var reverseVowels = function (s) {
 6     // corner case
 7     if (s === null || s.length === 0) return '';
 8 
 9     // normal case
10     let res = s.split('');
11     let vowels = new Set(['a', 'e', 'i', 'o', 'u', 'A', 'E', 'I', 'O', 'U']);
12     let left = 0;
13     let right = s.length - 1;
14     while (left < right) {
15         while (left < right && !vowels.has(res[left])) {
16             left++;
17         }
18         while (left < right && !vowels.has(res[right])) {
19             right--;
20         }
21         let temp = res[left];
22         res[left] = res[right];
23         res[right] = temp;
24         left++;
25         right--;
26     }
27     return res.join('');
28 };

Java实现

 1 class Solution {
 2     public String reverseVowels(String s) {
 3         // corner case
 4         if (s.length() == 0 || s == null) {
 5             return s;
 6         }
 7 
 8         // normal case
 9         String vowels = "aeiouAEIOU";
10         int left = 0;
11         int right = s.length() - 1;
12         char[] words = s.toCharArray();
13         while (left < right) {
14             while (left < right && vowels.indexOf(words[left]) == -1) {
15                 left++;
16             }
17             while (left < right && vowels.indexOf(words[right]) == -1) {
18                 right--;
19             }
20 
21             // swap
22             char temp = words[left];
23             words[left] = words[right];
24             words[right] = temp;
25             left++;
26             right--;
27         }
28         return new String(words);
29     }
30 }