题目来源
反转元音字母，我一开始想了想，记录下所有元音字母的位置，然后再反转。代码如下：

class Solution {
public:
    string reverseVowels(string s) {
        int n = s.size();
        vector<char> vowels{'a', 'e', 'i', 'o', 'u', 'A', 'E', 'I', 'O', 'U'};
        vector<int> pos;
        for (int i=0; i<n; i++)
            for (int j=0; j<vowels.size(); j++)
                if (s[i] == vowels[j]) {
                    pos.push_back(i);
                    break;
                }
        string res(s);
        int m = pos.size();
        for (int i=0; i<m; i++) {
            res[pos[i]] = s[pos[m-1-i]];
        }
        return res;
    }
};

然后看了一下Tag，发现写着双指针，发现自己太傻逼了…
改了改，代码如下：

class Solution {
public:
    string reverseVowels(string s) {
        int n = s.size();
        vector<char> vowels{'a', 'e', 'i', 'o', 'u', 'A', 'E', 'I', 'O', 'U'};
        int p1 = 0, p2 = n - 1;
        string res(s);
        while (p1 <= p2) {
            bool left = false, right = false;
            for (int i=0; i<vowels.size(); i++) {
                if (res[p1] == vowels[i]) {
                    left = true;
                }
                if (res[p2] == vowels[i]) {
                    right = true;
                }
            }
            if (left && right) {
                iter_swap(res.begin()+p1, res.begin()+p2);
                p1++;
                p2--;
            }
            if (!left)
                p1++;
            if (!right)
                p2--;
        }
        return res;
    }
};

可以AC，但是自己写的代码有点乱有点长。然后看了下别人的，代码如下：

class Solution {
public:
    string reverseVowels(string s) {
        int dict[256] = {0};
        dict['a'] = 1, dict['A'] = 1;
        dict['e'] = 1, dict['E'] = 1;
        dict['i'] = 1, dict['I'] = 1;
        dict['o'] = 1, dict['O'] = 1;
        dict['u'] = 1, dict['U'] = 1;
        int start = 0, end = (int)s.size() - 1;
        while(start < end){
            while(start < end && dict[s[start]] == 0) 
                start++;
            while(start < end && dict[s[end]] == 0) 
                end--;
            swap(s[start], s[end]);
            start++;
            end--;
        }
        return s;
    }
};