345. Reverse Vowels of a String

https://leetcode.com/problems/reverse-vowels-of-a-string/

本题大意：实现字符串中元音字母的反转。

解题思路：该题多种思路，这里给出两种。

1、用栈存储扫描到的元音字母，然后重新扫描，把栈中的元音字母填入字符串。时间复杂度o(2n)。

代码实现：

class Solution {
public:
    string reverseVowels(string s) {
        int n = s.length();
        stack<char> letter;
        for(int i = 0; i < n; i++)
        {
            if(s[i] == 'a' ||s[i] == 'e' || s[i] == 'i' || s[i] == 'o'||s[i] == 'u' || s[i] == 'A' || s[i] == 'E' || s[i] == 'I'||s[i] == 'O'||s[i] == 'U')
            {
                letter.push(s[i]);
            }
        }
        for(int i = 0; i < n; i++)
        {
            if(s[i] == 'a' ||s[i] == 'e' || s[i] == 'i' || s[i] == 'o'||s[i] == 'u' || s[i] == 'A' || s[i] == 'E' || s[i] == 'I'||s[i] == 'O'||s[i] == 'U')
            {
                s[i] = letter.top();
                letter.pop();
            }
        }
        return s;
    }
};

2、优化代码。从前往后扫描找到元音字母，再从后往前扫描找到元音字母，将两者对调。虽说是两个for循环嵌套起来，但是时间复杂度降低为o(n)了。

class Solution {
public:
    string reverseVowels(string s) {
        int n = s.length();
        int flag = n;
        for(int i = 0; i < flag; i++)
        {
            if(s[i] == 'a' ||s[i] == 'e' || s[i] == 'i' || s[i] == 'o'||s[i] == 'u' || s[i] == 'A' || s[i] == 'E' || s[i] == 'I'||s[i] == 'O'||s[i] == 'U')
            {
                for(int j = flag-1; j > i; j--)
                {
                    if(s[j] == 'a' ||s[j] == 'e' || s[j] == 'i' || s[j] == 'o'||s[j] == 'u' || s[j] == 'A' || s[j] == 'E' || s[j] == 'I'||s[j] == 'O'||s[j] == 'U')
                    {
                        char tmp = s[j];
                        s[j] = s[i];
                        s[i] = tmp;
                        flag = j;
                        break;
                    }
                }
            }
        }
        return s;
    }
};