[LeetCode] 824. Goat Latin

Description

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.

Analyse

  • 如果单词以元音(a,e,i,o,u)开头,在这个单词末尾增加"ma"

    "apple" -> "applema"

  • 如果单词以辅音开头,将单词第一个字母移动到末尾,然后在单词结尾增加"ma"

    "goat" -> "oatg" -> "oatgma"

  • 对每个单词,增加单词的序号个"a"到单词的末尾,序号从1开始

    "a b c" -> "ama bma cma" -> "amaa bmaaa cmaaaa"

简单题,直接上代码

string toGoatLatin(string S)
{
    stringstream ss(S);
    string tmp;
    string result;
    int index = 1;

    while (ss >> tmp)
    {
        if (tmp[0] == 'a' || tmp[0] == 'A' ||
            tmp[0] == 'e' || tmp[0] == 'E'    ||
            tmp[0] == 'i' || tmp[0] == 'I'    ||
            tmp[0] == 'o' || tmp[0] == 'O'    ||
            tmp[0] == 'u' || tmp[0] == 'U')
        {
            tmp.append("ma");
        }
        else
        {
            string first = tmp.substr(0, 1);
            tmp.erase(0, 1);
            tmp.append(first + "ma");
        }

        if (index != 1) result += " ";
        result += (tmp + string(index, 'a'));
        index++;
    }

    return result;
}
posted @ 2019-07-29 22:20  arcsinW  阅读(112)  评论(0编辑  收藏  举报