java - 嵌套 for 循环在两个字符串之间迭代

Question

我想使用 for 循环遍历每个字符串并依次输出每个字符。

String a = "apple";
String b = "class";

for (int i = 0;  i < a.length() ; i++) { // - 1 because 0 = 1
    System.out.print(a.charAt(i));
    for (int j = 0; j < b.length(); j ++) {
        System.out.print(b.charAt(j));
    }
}

我正在努力解决内部循环问题。

目前我的输出如下:

AClasspClasspClasslClasseClass

但是，我希望实现以下目标:

acplpalses

扩展问题:

将一个字符串反向输出，而另一个字符串正常输出怎么样？

当前尝试:

for (int i = a.length() - 1; i >= 0; i--) {
    System.out.println(a.charAt(i));
    for (int j = 0; j < b.length(); j ++) {
        System.out.println(b.charAt(j));
    }
}

但是，这只是输出如上，只是“Apple”以相反的顺序，格式与之前相同:

eclasslclasspclasspclassaclass

Answer 1

您不需要 2 个循环，因为您对两个字符串采用相同的索引

<小时/>

相同订单:

1. 简单的相同尺寸案例:

   for (int i = 0; i < a.length(); i++) {
       System.out.print(a.charAt(i));
       System.out.print(b.charAt(i));
   }
   

2. 复杂不同尺寸的案例:

   int minLength = Math.min(a.length(), b.length());
   for (int i = 0; i < minLength; i++) {
       System.out.print(a.charAt(i));
       System.out.print(b.charAt(i));
   }
   System.out.print(a.substring(minLength)); // prints the remaining if 'a' is longer
   System.out.print(b.substring(minLength)); // prints the remaining if 'b' is longer
   

<小时/>

顺序不同:

1. 简单的相同尺寸案例:

   for (int i = 0; i < a.length(); i++) {
       System.out.print(a.charAt(i));
       System.out.print(b.charAt(b.length() - i - 1));
   }
   

2. 复杂不同尺寸的案例:

   int minLength = Math.min(a.length(), b.length());
   for (int i = 0; i < minLength; i++) {
       System.out.print(a.charAt(i));
       System.out.print(b.charAt(b.length() - i - 1));
   }
   System.out.print(a.substring(minLength));
   System.out.print(new StringBuilder(b).reverse().substring(minLength));