Java-如何将字符串转换为piglatin？
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Java-How to translate a String to piglatin?
提问by T3rm3nator
I am writing a pig latin code and I am tripped up by how to get my program to identify where the next vowel in the word is if the first letter in the word is a consonant. Then it moves the first part of the word, up to the first vowel, to the end of the word and prints ay along with it. (ex. trees = eestray)
我正在编写一个猪拉丁代码，如果单词中的第一个字母是辅音，我被如何让我的程序识别单词中下一个元音的位置绊倒了。然后它移动单词的第一部分，直到第一个元音，移动到单词的末尾并打印 ay。（例如树木 = estray）
This is the code I have now
这是我现在的代码
// word is bringing in the string entered from the user
public static void translate(String word) {
String wordCap, pigLatin = "";
char vowels;
int lowest = 0, tempOne, tempTwo, tempThree, tempFour, tempFive;
wordCap = word.toUpperCase();
vowels = wordCap.charAt(0);
if (vowels == 'A' || vowels == 'E' || vowels == 'I' || vowels == 'O' || vowels == 'U') {
word = word + "way";
System.out.println(word);
}
else {
tempOne = wordCap.indexOf('A', 1);
if (lowest > tempOne && tempOne != -1) {
lowest = tempOne;
}
tempTwo = wordCap.indexOf('E', 1);
if (lowest > tempTwo && tempTwo != -1) {
lowest = tempTwo;
}
tempThree = wordCap.indexOf('I', 1);
if (lowest > tempThree && tempThree != -1) {
lowest = tempThree;
}
tempFour = wordCap.indexOf('O', 1);
if (lowest > tempFour && tempFour != -1) {
lowest = tempFour;
}
tempFive = wordCap.indexOf('U', 1);
if (lowest > tempFive && tempFive != -1) {
lowest = tempFive;
}
public static char vowel(String word) {
int start= 0, end= 0;
char vowels;
for (int i = 0; i < word.length(); i++) {
vowels = word.charAt(i);
if (vowels == 'A' || vowels == 'E' || vowels == 'I' || vowels == 'O' || vowels == 'U') {
end = i;
break;
}
}
return (char) end;
}
(in translate method)
（在翻译方法中）
for (int i = 0; i<wordCap.length(); i++) {
if (vowel(wordCap.charAt(i))) {
vowels = wordCap.charAt(i);
}
}
The problem now is that the vowel method is not an applicable method type. It says it must be a char?
现在的问题是元音方法不是适用的方法类型。它说它必须是一个字符？
回答by MeetTitan
Let me try to shorten that method for you ;)
让我尝试为您缩短该方法；)
Try something like this:
尝试这样的事情：
private static final char[] vowels = {'a', 'e', 'i', 'o', 'u'};
public static String translate(String word) {
int start = 0; // start index of word
int firstVowel = 0;
int end = word.length(); // end index of word
for(int i = 0; i < end; i++) { // loop over length of word
char c = Character.toLowerCase(word.charAt(i)); // char of word at i, lower cased
if(Arrays.asList(vowels).contains(c)) { // convert vowels to a list so we can use List.contains() convenience method.
firstVowel = i;
break; // stop looping
}
}
if(start != firstVowel) { // if start is not equal to firstVowel, we caught a vowel.
String startString = word.substring(firstVowel, end);
String endString = word.substring(start, firstVowel) + "ay";
return startString+endString;
}
return word; //couldn't find a vowel, return original
}
What this snippet does, is iterate over every character in the word, storing the index of the first vowel in the firstVowelvariable. Then, we get every character from firstVowelto end; and store it in startString. Then, we get every character from startto firstVowel; add "ay", and store it in endString. Finally, we concatenate these strings together and return them, resulting in the desired output.
这段代码的作用是遍历单词中的每个字符，将第一个元音的索引存储在firstVowel变量中。然后，我们得到从firstVowel到的每个字符end；并将其存储在startString. 然后，我们得到从start到的每个字符firstVowel；添加“ay”，并将其存储在endString. 最后，我们将这些字符串连接在一起并返回它们，从而得到所需的输出。
We can test this with System.out.println(translate("trees"));
我们可以用 System.out.println(translate("trees"));
EDIT:Without array, as requested:
编辑：没有数组，按要求：
public static String translate(String word) {
char a = 'a';
char e = 'e';
char i = 'i';
char o = 'o';
char u = 'u';
int start = 0;
int firstVowel = 0;
int end = word.length();
for(int i = 0; i < end; i++) {
char c = Character.toLowerCase(word.charAt(i));
if(c == a || c == e || c == i || c == o || c == u) {
firstVowel = i;
break;
}
}
if(start != firstVowel) {
String startString = word.subString(firstVowel, end);
String endString = word.subString(start, firstVowel) + "ay";
return startString+endString;
}
return word;
}
As you can see, arrays shorten things up quite a bit!
如您所见，数组大大缩短了事情的时间！
If you're feeling pedantic about the Arrays.asList().contains()call, we could define our own:
如果您对Arrays.asList().contains()调用感到迂腐，我们可以定义自己的：
public static boolean containsChar(char[] lookIn, char lookFor) {
boolean doesContainChar = false;
for(char c : lookIn) {
if(doesContainChar = c == lookFor)
break;
}
return doesContainChar;
}
回答by Bethany Louise
You might want to use a forloop to iterate through the letters of each word until it finds a vowel. Example:
您可能希望使用for循环遍历每个单词的字母，直到找到一个元音为止。例子：
String wordCap = word.toUpperCase();
char vowels;
for (int i=0; i<wordCap.length(); i++) {
if (isVowel(wordCap.charAt(i))) {
vowels = wordCap.charAt(i);
break;
}
}
Of course, I only used isVowel()for the sake of keeping the example concise. You'll have to identify it as a vowel the same way you did in your first ifstatement (or write an isVowel()method yourself).
当然，我使用只是isVowel()为了保持示例简洁。您必须像在第一个if语句中所做的那样（或isVowel()自己编写一个方法）将其识别为元音。
For modifying the word, you'll also want to declare a variable to hold the index of the vowel. The previous section of code could be added to for this, like so:
为了修改单词，您还需要声明一个变量来保存元音的索引。可以为此添加上一部分代码，如下所示：
String wordCap = word.toUpperCase();
char vowels;
int vowelIndex;
String newWord;
for (int i=0; i<wordCap.length(); i++) {
if (isVowel(wordCap.charAt(i))) {
vowels = wordCap.charAt(i);
vowelIndex = i;
break;
}
}
Then you could reference vowelIndexwhen modifying the word.
那么你可以vowelIndex在修改单词时参考。
if (vowelIndex == 0) {
newWord = word + "way";
} else {
newWord = word.substring(vowelIndex) + word.substring(0, vowelIndex) + "ay";
}
return word;
