Write a function that takes a string as input and reverse only the vowels of a string.
Example 1:
Given s = "hello", return "holle".
Example 2:
Given s = "leetcode", return "leotcede".
这道题让我们翻转字符串中的元音字母,元音字母有五个a,e,i,o,u,需要注意的是大写的也算,所以总共有十个字母。我们写一个isVowel的函数来判断当前字符是否为元音字母,如果两边都是元音字母,那么我们交换,如果左边的不是,向右移动一位,如果右边的不是,则向左移动一位,参见代码如下:
解法一:
class Solution {
public:
string reverseVowels(string s) {
int left = , right= s.size() - ;
while (left < right) {
if (isVowel(s[left]) && isVowel(s[right])) {
swap(s[left++], s[right--]);
} else if (isVowel(s[left])) {
--right;
} else {
++left;
}
}
return s;
}
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';
}
};
或者我们也可以用自带函数find_first_of和find_last_of来找出包含给定字符串中任意一个字符的下一个位置进行交换即可:
解法二:
class Solution {
public:
string reverseVowels(string s) {
int left = , right = s.size() - ;
while (left < right) {
left = s.find_first_of("aeiouAEIOU", left);
right = s.find_last_of("aeiouAEIOU", right);
if (left < right) {
swap(s[left++], s[right--]);
}
}
return s;
}
};
我们也可以把元音字母都存在一个字符串里,然后每遇到一个字符,就到元音字符串里去找,如果存在就说明当前字符是元音字符,参见代码如下:
解法三:
class Solution {
public:
string reverseVowels(string s) {
int left = , right = s.size() - ;
string t = "aeiouAEIOU";
while (left < right) {
if (t.find(s[left]) == string::npos) ++left;
else if (t.find(s[right]) == string::npos) --right;
else swap(s[left++], s[right--]);
}
return s;
}
};
类似题目:
参考资料:
https://leetcode.com/discuss/99048/easy-to-understand-c-solution
https://leetcode.com/discuss/99047/super-clean-solution-using-find_first_of-and-find_last_of
https://leetcode.com/discuss/99062/java-two-pointers-solution-easy-understand-finish-interview