LeetCode 824. Goat Latin

最新推荐文章于 2020-05-17 12:14:32 发布

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本文介绍了一个将英语句子转换为Goat Latin形式的C++实现方案。Goat Latin是一种模仿Pig Latin的游戏语言，通过特定规则修改每个单词，并根据单词在句子中的位置增加字母'a'。

题目链接：https://leetcode.com/problems/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.

题目解析：对每个字符进行处理，标记前一个位置是否为空格，如果是开头第一个字符就单独处理，记录非元音字符即可。在空格和最后塞入后缀，如果有非元音开头字符的话，先加上该字符。

代码如下：0ms Accepted beating 100%

class Solution {
public:
    string toGoatLatin(string S) {
        string ans;
        string ss = "maa";
        char c = '\0';
        bool isBlank = true;
        for (int i = 0; i <= S.size(); i++)
        {
            char cur = S[i];
            if (cur == ' ' || i == S.size())
            {
                if (c != '\0')
                {
                    ans.push_back(c);
                    c = '\0';
                }
                ans.append(ss);
                if (cur == ' ')
                {
                    ans.push_back(' ');
                    isBlank = true;
                }
                ss += "a";
            }
            else
            {
                if (isBlank && !(cur == 'a' || cur == 'A' || cur == 'e' || cur == 'E' || cur == 'i' || cur == 'I' || cur == 'o' || cur == 'O' || cur == 'u' || cur == 'U'))
                {
                    c = cur;
                    isBlank = false;
                    continue;
                }
                isBlank = false;
                ans.push_back(cur);
            }
        }
        return ans;
    }
};