Question

我有一个java程序，其中以下是我想要实现的目标：

first input: ABC

second input: xyz

output: AxByCz

我的Java程序如下：

    import java.io.*;

    class DisplayStringAlternately 
    {
        public static void main(String[] arguments)
        {
            String firstC[], secondC[];

            firstC = new String[] {"A","B","C"};
            secondC = new String[] {"x","y","z"};

            displayStringAlternately(firstC, secondC);        
        }

        public static void displayStringAlternately (String[] firstString, String[] secondString)
        {
           int combinedLengthOfStrings = firstString.length + secondString.length;

           for(int counter = 1, i = 0; i < combinedLengthOfStrings; counter++, i++)
           {
               if(counter % 2 == 0)
               {
                   System.out.print(secondString[i]);
               }
               else 
               {
                   System.out.print(firstString[i]);
               }
           }
        }        
    }

但是我遇到了以下运行时错误：

    Exception in thread "main" java.lang.ArrayIndexOutOfBoundsException: 3
    AyC at DisplayStringAlternately.displayStringAlternately(DisplayStringAlternately.java:23)
        at DisplayStringAlternately.main(DisplayStringAlternately.java:12)
    Java Result: 1

我的Java程序有什么错误？

Answer 1

正如您所评论的，如果两个数组长度相同，您只需执行

即可

 firstC = new String[] {"A","B","C"};
 secondC = new String[] {"x","y","z"};

然后

          for(int i = 0; i < firstC.length; i++)  {
                   System.out.print(firstC[i]);
                   System.out.print(secondC[i]);
           }

Answer 2

如果两个数组具有相同的长度for，则应在i < anyArray.length时继续循环。

此外，您不需要任何counter来确定应首先打印哪个阵列。只需硬编码将从firstString打印第一个元素，然后从secondString打印下一个元素。

因此，您的displayStringAlternately方法可能看起来像

public static void displayStringAlternately(String[] firstString,
        String[] secondString) {
    for (int i = 0; i < firstString.length; i++) {
        System.out.print(firstString[i]);
        System.out.print(secondString[i]);
    }
}

无论如何，你的代码抛出ArrayIndexOutOfBoundsException，因为每次你决定从哪个数组打印元素递增i时，你都会以这种方式跳过数组

 i=0     i=2
{"A","B","C"};

{"x","y","z"};    
     i=1     i=3 
             ^^^-here is the problem

所以你看到你的代码试图从第二个数组中访问不在其中的元素（它超出了它的范围）。

Answer 3

使用字符串的组合长度是错误的，例如，当{i> = secondString.length时，secondString[i]会导致异常。

Answer 4

尝试使用高性能的以下工作代码

public static void main(String[] arguments)
{
    String firstC[], secondC[];

    firstC = new String[] {"A","B","C"};
    secondC = new String[] {"x","y","z"};
    StringBuilder  builder = new StringBuilder();
    for (int i = 0; i < firstC.length; i++) {
        builder.append(firstC[i]);
        builder.append(secondC[i]);
    }
    System.out.println(builder.toString()); 
}

Answer 5

public class concad {

public void main(String[] args) {

    String s1 = "RAMESH";
    String s2 = "SURESH";

    int i;
    int j;

    for (i = 0; i < s1.length(); i++) {
        System.out.print(s1.charAt(i));
        for (j = i; j <= i; j++) {
            if (j == i) {
                System.out.print(s2.charAt(j));
            }

        }

    }
}

}

Answer 6

我已经提到了两个字符串。然后在内部for循环中传递一个计数器变量和第二个字符串，然后为每个偶数位置传递代码＆＃34;计数器％2＆＃34;。如果有任何问题，请检查然后在下面评论。

public class AlternatePosition {

public static void main(String[] arguments) {

    String abc = "abcd";
    String def = "efgh";
    displayStringAlternately(abc, def);
}

public static void displayStringAlternately(String firstString, String secondString) {
    for (int i = 0; i < firstString.length(); i++) {
        for (int counter = 1, j = 0; j < secondString.length(); counter++, j++) {
            if (counter % 2 == 0) {
                System.out.print(secondString.charAt(i));
                break;
            } else {
                System.out.print(firstString.charAt(i));
            }
        }
    }
}
}

在Java中交替显示2个字符串

6 个答案: