Given a 0-indexed string s, permute s to get a new string t such that:
i with 0 <= i < s.length such that s[i] is a consonant, then t[i] = s[i].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 <= 105s consists only of letters of the English alphabet in uppercase and lowercase.class Solution {
public String sortVowels(String s) {
char[] cs = s.toCharArray();
Set<Character> vowels = new HashSet<>(Arrays.asList('a', 'e', 'i', 'o', 'u', 'A', 'E', 'I', 'O', 'U'));
List<Character> l = new ArrayList<>();
for (char c : cs)
if (vowels.contains(c))
l.add(c);
Collections.sort(l);
StringBuilder sb = new StringBuilder();
int vi = 0;
for (char c : cs)
sb.append(vowels.contains(c) ? l.get(vi++) : c);
return sb.toString();
}
}