java - 在某个字符第一次出现时停止

Question

我只想删除“brealdeke”中的“a”这个程序有效，它打印“breldeke ”，但如果我把“brealdeake”在该字符串中带有 2 个“a”，它会变得狂暴并打印:breldeakebrealdeke 如何修复它？谢谢我真的希望它看起来像这样:

class Example {
    public static String suppression(char c, String s) {
        for (int i = 0; i < s.length(); i++) {
            if (s.charAt(i) == c) {
                int position = i;
                for (int a = 0; a < position; a++) {
                    System.out.print(s.charAt(a));
                }
                for (int b = position + 1; b < s.length(); b++) {
                    System.out.print(s.charAt(b));
                }
            }

        }
        return "";
    }

    public static void main(String[] args) {
        // prints "breldeke"
        System.out.println(suppression('a', "brealdeke"));
        // prints "breldeakebrealdeke"
        System.out.print(suppression('a', "brealdeake"));
    }
}

Answer 1

你可以尝试:

"banana".replaceFirst("a", "");

这会返回bnana

编辑:希望这不包括您尚未学到的任何内容

public static void main(String[] args) {
    String word = "banana";
    String strippedWord = "";
    boolean found = false;
    for (int i = 0; i < word.length(); i++) {
        if (word.charAt(i) == 'a' && !found) found = !found;
        else strippedWord += word.charAt(i);
    }
    System.out.println(strippedWord);
}

这将打印 bnana

EDIT2:你说你想在一个函数中使用它，这同样适用:

public static String suppression(char c, String word) {
    String strippedWord = "";
    boolean charRemoved = false;    // This is a boolean variable used to know when the char was skipped!
    for (int i = 0; i < word.length(); i++) {
        // If the current letter is for example 'a' and we haven't yet skipped the char, skip this char we're at 
        if (word.charAt(i) == c && charRemoved == false) charRemoved = true;
        else strippedWord += word.charAt(i);
    }
    return strippedWord;
}

public static void main(String[] args) {
    // prints "breldeke"
    System.out.println(suppression('a', "brealdeke"));
    // prints "breldeake"
    System.out.print(suppression('a', "brealdeake"));
}