力扣每日一题2022-04-21简单题：山羊拉丁文

山羊拉丁文

• • 题目描述
  • 思路
  • • 模拟
    • • Python实现
      • Java实现

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题目描述

山羊拉丁文

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思路 模拟

遍历字符串，分割找到每个单词，依照题意进行 *** 作即可。

Python实现

VOWELS = {"a", "e", "i", "o", "u", "A", "E", "I", "O", "U"}

class Solution:
    def toGoatLatin(self, sentence: str) -> str:
        n = len(sentence)
        i, cnt = 0, 1
        words = list()
        while i < n:
            j = i
            while j < n and sentence[j] != " ":
                j += 1
            cnt += 1
            if sentence[i] in VOWELS:
                words.append(sentence[i:j] + "m" + "a" * cnt)
            else:
                words.append(sentence[i+1:j] + sentence[i] + "m" + "a" * cnt)
            i = j + 1
        
        return " ".join(words)

Java实现

class Solution {
    private static final Set<Character> VOWELS = new HashSet<>();

    static {
        VOWELS.add('a');
        VOWELS.add('e');
        VOWELS.add('i');
        VOWELS.add('o');
        VOWELS.add('u');
        VOWELS.add('A');
        VOWELS.add('E');
        VOWELS.add('I');
        VOWELS.add('O');
        VOWELS.add('U');
    }

    public String toGoatLatin(String sentence) {
        StringBuilder sb = new StringBuilder();
        int n = sentence.length(), cnt = 1;
        for (int i = 0; i < n; i++) {
            char ch = sentence.charAt(i++);
            if (VOWELS.contains(ch)) {
                sb.append(ch);
            }
            while (i < n && sentence.charAt(i) != ' ') {
                sb.append(sentence.charAt(i++));
            }
            if (!VOWELS.contains(ch)) {
                sb.append(ch);
            }
            sb.append("m");
            for (int j = 0; j <= cnt; j++) {
                sb.append("a");
            }
            if (i < n-1) {
                sb.append(" ");
            }
            cnt++;
        }
        return sb.toString();
    }
}