Question

package num20;

import TurtleGraphics.KeyboardReader;

public class Num20main {

public static void main(String[] args) {
    KeyboardReader reader = new KeyboardReader();
    System.out.println("Enter String");
    String sentence = reader.readLine();
    StringBuilder sb = new StringBuilder(sentence);
    System.out.println(sb.toString());
    //String e = "egg";
    for(int x = 0; x < sentence.length(); x++){
        String e = "egg";
        char l = sb.charAt(x);
        int index = x;
        if(l == ('a') || l == ('A') || l == ('e') || l == ('E') || l == ('I') || l == ('i') || l == ('O') || l == ('o') || l == ('U') || l == ('u')){
            sb.insert(index, e);
            System.out.println(sb.toString());
        }
    }
    System.out.println(sb.toString());

}

}

此代码打印出来：

输入字符串

我爱Java

我爱Java

eggI Love Java

eggeggI Love Java

eggeggeggI Love Java

eggeggeggeggI Love Java

eggeggeggeggI Love Java

它应该打印出“eggI Leggovegge Jeggavegga”

Answer 1

您正在添加String的开头。所以发生了什么，你添加“鸡蛋”，但你的索引不会前进。 “我喜欢java”'我'在索引0。但是当你添加一个“鸡蛋”时，我们就有了“eggI love java”。现在'我'在索引3处，但此时你的x是1。

最简单的方法是以相反的方式运行：

for(int x = sentence.length()-1 ; x <= 0 ; x--){ //The content of your for here }

这样做，添加“鸡蛋”不会影响您的功能。

Answer 2

尝试这种方式：

    String sentence = "I love Java";
    System.out.println(Arrays.stream(sentence.split(" ")).map(str -> str.replaceAll("(?i)[aeiou]", "fruit$0")).collect(Collectors.joining(" ")));

基本上，您首先进行不区分大小写的查找，然后将匹配替换为匹配+＆＃34;水果＆＃34;。

Answer 3

您的索引以与x相同的速度移动，但每次替换时，您插入3而不是1个字符。

我使用了一个固定的String来进行简单的测试，并且为了便于控制，最后添加了预期的结果：

public class Num20 {

    public static void main(String[] args) {
        String sentence = "I Love Java";
        StringBuilder sb = new StringBuilder(sentence);
        System.out.println(sb.toString());
        // String e = "egg"; Good idea to declare this unchanged variable
        // here, but let us call it 'egg'. 
        String egg = "egg";
        int index = 0; // index out of loop 
        for (int x = 0; x < sentence.length(); x++){
            char l = sb.charAt (index); // sb is under constant change, 
            // we need to insert at a the place of sb, not sentence
            // hence we use the faster moving index 'index', not 'x'
//            if (l == ('a') || l == ('A') || l == ('e') || l == ('E') || l == ('I') || l == ('i') || l == ('O') || l == ('o') || l == ('U') || l == ('u')) {
            // not wrong, but more brief:
            if ("aeiouAEIOU".indexOf (l) != -1) {
                sb.insert (index, egg);
                System.out.println(sb.toString());
                // we inserted 3 characters before the vowel 
                index +=3;
            }
            // we have to forward one (more) step for every x in sentence, too:
            ++index;
        }
        System.out.println(sb.toString());
        System.out.println("eggI Leggovegge Jeggavegga?");
    }
}

无法在另一个字符串

3 个答案: