Question

我开始这是因为我完全无聊但是由于这个错误，我一直坐在这里很久以来最后决定把它带到stackOverFlow。这是我写的代码。

我试图通过跳过1个索引来打印字符。但是当有重复时，我想打印一个空格，这会使大字符串中的单词不同。

更新的问题：一切都已修复，除了我不能增加I值超过1.我在下面的程序中评论它。请看一下。

让我切断追逐并达到目的。我需要这个输出＆＃34; Vishnu Vardhan＆＃34;来自这个字符串＆＃34; aVeIwSjHaNgUaaVdAgRjDkHxAmN＆＃34 ;;

我唯一的要求是，如果字符串有两个相同的字母，则必须打印空格。所以＆＃34; aVeIwSjHaNgU [aa] VdAgRjDkHxAmN＆＃34;括号中的aa必须用空格代替。

它必须是动态的，如果重复任何字符，它必须打印一个空格并跳转到所需的下一个字符并打印它。

这是更新的程序。使用其中一条评论的帮助。

公共阶层Decrypter {

public static void main(String[] args) {
    String Text = "aVeIwSjHaNgUaaVdAgRjDkHxAmN";
    char[] convertedText = Text.toCharArray();//converted string to char array

    for (int i = 1; i < convertedText.length; i++) { //Looping it to print alternate chars

        /* if the character at an index is same as the character at next index then
         add a space and increase index value by 2 so I can print the required char*/
        if (i + 1 < convertedText.length) {
            if (Text.charAt(i) == Text.charAt(i + 1)) {

                i++;// Increasing I value by 2 here will give me required output. Everything is perfect now

                System.out.printf("%s ", convertedText[i]);
            } else {
                System.out.printf("%s", convertedText[i]);
                i++;

            }
        }
    }
}

}

当前输出：VISHNUadgjkxm

必需的输出：VISHNU VARDHAN

Answer 1

我不知道是否需要将字符串转换为charArray，但我希望这样做。如果您有问题需要修改，请在下方发表评论。

    String text = "aVeIwSjjHaNgUkkVarqddlhxn";
    //this is the holder of your new processed text.
    String newText = "";

    //start of counter. it may start in any number depends on requirements.
    int x = 0;

    //loop while the x is lessthan the length of your string.
    while(x < text.length()){

        //check if the x + 1 is not equal to the length of your string to avoid StringIndexOutOfBoundsException
        if((x+1) != text.length()){

            //in this area it will check if the current char is the same on next index
            if(text.charAt(x) == text.charAt(x+1)){

                // this will concatenate/append the value of char with space 
                newText += text.charAt(x) +" ";
                // this will increase the value of your x by 1 and at the bottom there are also x++ that will add 1 to your x so the total sum of x if (text.charAt(x) == text.charAt(x+1)) are true, will be 2.
                x++;
            }

        }
        newText += text.charAt(x);
        x++;
    }
    System.out.println(newText);

输出：

aVeIwSj jHaNgUk kVarqd dlhxn

如果这不是您要找的，请更新您的问题。

Answer 2

修正：

/**
 *
 * @author Chintu
 */
public class Decrypter {

public static void main(String[] args) {
    String Text = "aVeIwSjHaNgUkkVarqdlhxn";
    char[] convertedText = Text.toCharArray();//converted string to char array

    for (int i = 1; i < convertedText.length; i++) { //Looping it to print alternate chars

        /* if the character at an index is same as the character at next index then
         add a space and increase index value by 2 so I can print the required char*/
        if (i+1 < convertedText.length) {
            if (Text.charAt(i) == Text.charAt(i + 1)) {
                i++;
                System.out.printf("%s ",convertedText[i]);
            }
        }
        System.out.printf("%s", convertedText[i]);
    }
}
}

Answer 3

固定

公共阶层Decrypter {

public static void main(String[] args) {
    String Text = "aVeIwSjHaNgUaaVdAgRjDkHxAmN";
    char[] convertedText = Text.toCharArray();//converted string to char array

    for (int i = 0; i < convertedText.length; i++) { //Looping it to print alternate chars

        /* if the character at an index is same as the character at next index then
         add a space and increase index value by 2 so I can print the required char*/
        if (i + 1 < convertedText.length) {
            if (Text.charAt(i) == Text.charAt(i + 1)) {

                i += 2;   
                System.out.printf(" %s", convertedText[i]);
            } else {
                i++;
                System.out.printf("%s", convertedText[i]);

            }
        }
    }
}

}

输出：VISHNU VARDHAN

比较两个字符后，指数值不会增加

3 个答案: