如何限制 while (iterator.hasNext()) 迭代？

Question

我正在 Java 使用 Generex 库，根据给定的正则表达式打印字符串。

有些R.Es可以生成无限字符串，我只想处理它们，但还不能。我的代码看起来像；

Generex generex = new Generex(regex);
Iterator iterator = generex.iterator();
    System.out.println("Possible strings against the given Regular Expression;\n");
    while (iterator.hasNext()) {
        System.out.print(iterator.next() + " ");
    }

如果我输入 (a)* 作为正则表达式，输出应该如下所示

a aa aaa aaaa aaaaa aaaaaa aaaaaaa aaaaaaaa aaaaaaaaa ...

如何限制该循环的结果？

Answer 1

假设您希望打印前 8 项，然后如果要打印更多项则添加 "..."。您可以按如下方式进行：

int limit = 8;
int current = 0;
while (iterator.hasNext()) {
    if (current != 0) {
        System.out.print(" ");
    }
    System.out.print(iterator.next());
    // If we reach the limit on the number of items that we print,
    // break out of the loop:
    if (++current == limit) {
        break;
    }
}
// When we exit the loop on break, iterator has more items to offer.
// In this case we should print an additional "..." at the end
if (iterator.hasNext()) {
    System.out.print(" ...");
}

Answer 2

在你的情况下，我认为字符串的长度比打印的元素数量重要得多，所以我认为以下解决方案可能更好：

Generex generex = new Generex(regex);
Iterator iterator = generex.iterator();
System.out.println("Possible strings against the given Regular Expression;\n");
StringBuilder sb = new StringBuilder();
int limitOfChars = 100; //for example
while (iterator.hasNext()) {
    String next = iterator.next();
    if (sb.length() + next.length() > limitOfChars) break;
    sb.append(next + " ");
}
System.out.println(sb.toString() + " ... ");

如何限制 while (iterator.hasNext()) 迭代？

How to Limit while (iterator.hasNext()) iterations?

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