如何在字符串中找到第一个元音的索引？ [英] how to find index of first vowel in a string?

问题描述

这部分我想实现，但不知道如何：
在Pig Latin中，所有辅音直到第一个元音应该移动到单词的结尾。所以当laptop仍然是aptoplay，string应该成为ingStray。

This part i want implemented but don't know how: In Pig Latin, all the consonants until the first vowel should be moved to the end of the word. So while "laptop" will still be "aptoplay", "string" should become "ingStray".

我有这个，第一部分工作很好。 >

I have this, the first part works just fine.

public static String doStuff(String word) {
    int number = 0;
    char[] vowel = { 'a', 'e', 'i', 'o', 'u' };
    char first = word.charAt(0);
    char second = Character.toLowerCase(first);
    if (second == 'a' || second == 'e' || second == 'i' || 
            second == 'o' || second == 'u') {
        word = word + "ay";
    } else {
        for (int i = 0; i < word.length(); i++) {
        for (int j = 0; j < 5; j++) {
            if (word.charAt(i) == vowel[j]) {
                word = word.substring(i + 1) + word.substring(i) + "ay";
                break;
            }
        }
    }
    return word;
}

推荐答案

code> break 应该突破两个循环。要这样做，在你的第一个循环之前放置一个标志

First of all, your break should break out of both loops. To do so, put a flag before your first loop like

outerloop:
for (int i = 0; i < word.length(); i++) {

; put break outerloop;

其次，

word = word.substring(i + 1) + word.substring(i) + "ay";

不工作，因为当您找到元音时， i + 1 应该只是 i 。然后，ay之前的子字符串的其余部分应该是开头的辅音，因此 0，i 而不是只是 i 。总之，这提供了

doesn't work, because when you find the vowel, you want to keep it in front, so i + 1 should be just i. Then, the rest of the substring before "ay" should be the consonants at the beginning, so 0, i instead of just i. In all, this gives

outerloop:
    for (int i = 0; i < word.length(); i++) {
        for (int j = 0; j < 5; j++) {
            if (word.charAt(i) == vowel[j]) {
                word = word.substring(i) + word.substring(0, i) + "ay";
                break outerloop;
            }
        }
    }

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