LeetCode 2785. Sort Vowels in a String

Given a 0-indexed string s, permute s to get a new string t such that:

• All consonants remain in their original places. More formally, if there is an index i with 0 <= i < s.length such that s[i] is a consonant, then t[i] = s[i].

• The vowels must be sorted in the nondecreasing order of their ASCII values. More formally, for pairs of indices i, j with 0 <= i < j < s.length such that s[i] and s[j] are vowels, then t[i] must not have a higher ASCII value than t[j].

Return the resulting string.

The vowels are 'a', 'e', 'i', 'o', and 'u', and they can appear in lowercase or uppercase. Consonants comprise all letters that are not vowels.

Example 1:

Input: s = "lEetcOde"

Output: "lEOtcede"

Explanation: 'E', 'O', and 'e' are the vowels in s; 'l', 't', 'c', and 'd' are all consonants. The vowels are sorted according to their ASCII values, and the consonants remain in the same places.

Example 2:

Input: s = "lYmpH"

Output: "lYmpH"

Explanation: There are no vowels in s (all characters in s are consonants), so we return "lYmpH".

Constraints:

• 1 <= s.length <= 105

• s consists only of letters of the English alphabet in uppercase and lowercase.

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给定一个索引为 0 的字符串 s，对 s 进行排列以获得新字符串 t，使得：

所有辅音都保留在原来的位置。 更正式地说，如果存在一个索引 i 且 0 <= i < s.length 且 s[i] 是辅音，则 t[i] = s[i]。

元音必须按照其 ASCII 值的非递减顺序排序。 更正式地说，对于索引 i、j 对，其中 0 <= i < j < s.length，使得 s[i] 和 s[j] 是元音，则 t[i] 的 ASCII 值不得高于 t[j]。

返回结果字符串。

元音为“a”、“e”、“i”、“o”和“u”，它们可以小写或大写形式出现。 辅音包括所有非元音的字母。

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下面是代码：

Runtime: 121 ms, faster than 40.00% of Java online submissions for Sort Vowels in a String.

Memory Usage: 45.3 MB, less than 100.00% of Java online submissions for Sort Vowels in a String.