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Java - 简单的英语到 leet 转换器

转载 作者:行者123 更新时间:2023-11-30 10:55:11 34 4
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我正在尝试创建一个简单的“英语到 leet”转换器,但我不知道要使用什么方法将字符串转换为 leet。

由于某些原因,当我有字符串 english[] 和 leet[] 时,我不能使用 equalsIgnoreCase

    public static void main(String[] args) {
String english[] = {"A", "B", "C", "D", "E", "F", "G", "H", "I", "J", "K", "L", "M", "N", "O", "P", "Q", "R", "S", "T", "U", "V", "W", "X", "Y", "Z"};
String leet[] = {"4", "8", "(", ")", "3", "}", "6", "#", "!", "]", "X", "|", "M,", "N", "0", "9", "Q", "2", "Z", "7", "M", "V", "W", "X", "J", "Z"};
String result = "";

Scanner sc = new Scanner(System.in);
String Str = sc.nextLine();

for (int i = 0 ; i < english.length ; i++) {

if (Str.equalsIgnoreCase(english)) {
// convert to leet
}
}
}

最佳答案

因为英语数组只是按顺序列出“A”到“Z”,你甚至需要循环遍历英语数组,只需这样做:

Scanner sc = new Scanner(System.in);
String Str = sc.nextLine().toUpperCase(); // convert all to upper case so that you don't need equalsIgnoreCase()
for (int i = 0; i < Str.length(); ++i) {
result +=leet[Str.charAt(i) - 'A']; // Str.charAt(i) - 'A' this will give you the correct index in leet
}

关于Java - 简单的英语到 leet 转换器,我们在Stack Overflow上找到一个类似的问题: https://stackoverflow.com/questions/33416987/

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