A sentence S is given, composed of words separated by spaces. Each word consists of lowercase and uppercase letters only.

We would like to convert the sentence to “Goat Latin” (a made-up language similar to Pig Latin.)

The rules of Goat Latin are as follows:

If a word begins with a vowel (a, e, i, o, or u), append "ma" to the end of the word. For example, the word ‘apple’ becomes ‘applema’.

If a word begins with a consonant (i.e. not a vowel), remove the first letter and append it to the end, then add "ma". For example, the word "goat" becomes "oatgma".

Add one letter 'a' to the end of each word per its word index in the sentence, starting with 1. For example, the first word gets "a" added to the end, the second word gets "aa" added to the end and so on. Return the final sentence representing the conversion from S to Goat Latin.

Example 1:

Input: "I speak Goat Latin"
Output: "Imaa peaksmaaa oatGmaaaa atinLmaaaaa"

Example 2:

Input: "The quick brown fox jumped over the lazy dog"
Output: "heTmaa uickqmaaa rownbmaaaa oxfmaaaaa umpedjmaaaaaa overmaaaaaaa hetmaaaaaaaa azylmaaaaaaaaa ogdmaaaaaaaaaa"

Notes:

S contains only uppercase, lowercase and spaces. Exactly one space between each word.
1 <= S.length <= 150.

Solution

Time complexity : O(n)
Space complexity : O(n)

class Solution {
public:
    string toGoatLatin(string S) {
        const string vowels = "AEIOUaeiou";
        int cnt = 0;
        string word, ans;
        istringstream iss(S);
        while (iss >> word) {
            if (vowels.find(word[0]) == string::npos)
                word = word.substr(1) + word[0];
            ans += " " + word + "ma" + string(++cnt, 'a');
        }
        return ans.substr(1);
    }
};

使用istringstream歷遍S中的每個word。
npos指的是字串的尾巴，找不到時會跑到這邊。
一個一個加到ans中，最後ans頭會多出一個空白，跳過它即可。