345. Reverse Vowels of a String (Easy)

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".

Note:

The vowels does not include the letter "y".

two pointers

Solution 1:

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：

string reverseVowels(string s) {
    int left = 0, right= s.size() - 1;
    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';
}

Solution 2:

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：

(13 ms)

string reverseVowels(string s) { 
    int front = 0, back = s.size()-1;
    while (front < back) {
        // a e i o u
        while (front < back && s[front] != 'a' && s[front] != 'e' && s[front] != 'i' && s[front] != 'o' && s[front] != 'u' && s[front] != 'A' && s[front] != 'E' && s[front] != 'I' && s[front] != 'O' && s[front] != 'U') ++front;
        while (front < back && s[back] != 'a' && s[back] != 'e' && s[back] != 'i' && s[back] != 'o' && s[back] != 'u' && s[back] != 'A' && s[back] != 'E' && s[back] != 'I' && s[back] != 'O' && s[back] != 'U') --back;

        if (front < back) swap(s[front++], s[back--]);
    }
    return s;
}

(fastest version 9ms)

string reverseVowels(string s) {
    int left = 0, right = s.size() - 1;
    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;
}

Solution 3: (slowest 16ms)

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：

string reverseVowels(string s) {
    int left = 0, right = s.size() - 1;
    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;
}