java HashMap get方法看起来像是在返回一个副本，对吗？

Question

我有一个程序，可以逐字读取包含小故事的文本文件。检查每个单词是否有“<；名词>；”形式的标签，“<；动词>；”或者类似的东西。如果找到标签，它将使用与标签对应的ArrayList中的随机字替换该标签，方法如下：

private String getSubstitute(String label) {
        if (listMap.get(label) != null){
            String substitute = randomFrom(listMap.get(label));
            listMap.get(label).remove(substitute);
            return randomFrom(listMap.get(label));
        }
        if (label.equals("number")){
            return " " + myRandom.nextInt(50) + 5;
        }
    return "**UNKNOWN**";
}

这是相当好的工作，除了在每一个单词的使用，它应该从它的相应列表中删除，这样它就不能被重复使用。我试着在上面代码的第4行中这样做。HashMaplistMap<String, ArrayList>是一个实例变量，所以我觉得调用 listMap.get(label).remove(substitute)，我将直接访问listMap中包含的ArrayList并将其删除。但是，当我运行代码时，有时会看到以前使用过的单词重复出现。我错过了什么

举个有趣的例子，下面是在一个实例中打印出来的内容：

This is a slippery story about how a blue tiger became a blue tiger. Once upon a time, about 295 decades ago, angry, angry pangolins roamed the earth. One of them was named Jermaine. This pangolin was alone in the world. Then it became a orange pangolin living in Ecuador. This animal loved to think and surrender. In the morning it would eat a green fig, and later eat a angry mango for a snack.

这是这个故事产生的文本文件：

This is a <adjective> story about how a <color> <animal> became a <color> <animal>. Once upon a time, about <number> <timeframe> ago, <adjective>, <adjective> <animal>s roamed the earth. One of them was named <name>. This <animal> was alone in the world. Then it became a <color> <animal> living in <country>. This animal loved to <verb> and <verb>. In the morning it would eat a <color> <fruit>, and later eat a <adjective> <fruit> for a snack.

Answer 1

问题是，您肯定会删除分配给substitute的那个。但你不能退回那个。你可以随机返回另一个。所以你没有移除你返回的那个

应该是的

    private String getSubstitute(String label) {
        if (listMap.get(label) != null){
            String substitute = randomFrom(listMap.get(label));
            listMap.get(label).remove(substitute);
            return substitute;
        }
        if (label.equals("number")){
            return " " + myRandom.nextInt(50) + 5;
        }
        return "**UNKNOWN**";
     }

Answer 2

你不能删除getSubstitute返回的内容。您将返回最后生成的下一个随机单词：

return randomFrom(listMap.get(label));

Answer 3

5号线

return substitute;

而不是

return randomFrom(listMap.get(label));