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.

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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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給定一個索引為 0 的字符串 s，對 s 進(jìn)行排列以獲得新字符串 t，使得：

所有輔音都保留在原來的位置。 更正式地說，如果存在一個索引 i 且 0 <= i < s.length 且 s[i] 是輔音，則 t[i] = s[i]。

元音必須按照其 ASCII 值的非遞減順序排序。 更正式地說，對于索引 i、j 對，其中 0 <= i < j < s.length，使得 s[i] 和 s[j] 是元音，則 t[i] 的 ASCII 值不得高于 t[j]。

返回結(jié)果字符串。

元音為“a”、“e”、“i”、“o”和“u”，它們可以小寫或大寫形式出現(xiàn)。 輔音包括所有非元音的字母。

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下面是代碼：

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.