Write a function to reverse only the vowels of a string. Example: Given s = “hello”, return “holle”.
A vowel is one of the following 10 letters (both uppercase and lowercase).
1 2 3 4 | inline bool isVowel(char c) { return c == 'a' || c == 'e' || c == 'i' || c == 'o' || c == 'u' || c == 'A' || c == 'E' || c == 'I' || c == 'O' || c == 'U'; } |
inline bool isVowel(char c) {
return c == 'a' || c == 'e' || c == 'i' || c == 'o' || c == 'u' ||
c == 'A' || c == 'E' || c == 'I' || c == 'O' || c == 'U';
}To swap only the vowels we need two index pointers, pointing to the start and the end of the string, only stopping at the vowels until they meet in the middle.
1 2 3 4 5 6 7 8 9 10 11 12 13 14 | string reverseVowels(string s) { int i = 0; int j = s.length() - 1; while (i < j) { while (i < j && !isVowel(s[i])) { i ++; } while (i < j && !isVowel(s[j])) { j --; } swap(s[i++], s[j--]); } return s; } |
string reverseVowels(string s) {
int i = 0;
int j = s.length() - 1;
while (i < j) {
while (i < j && !isVowel(s[i])) {
i ++;
}
while (i < j && !isVowel(s[j])) {
j --;
}
swap(s[i++], s[j--]);
}
return s;
}O(n) time complexity and O(1) space complexity.
–EOF (The Ultimate Computing & Technology Blog) —
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