Java将Arraylist分割成更小的ArrayList

Question

我有一个大小为258的ArrayList

现在我希望将其拆分为三个不同的 ArrayList，为此我创建了以下代码:

    Integer start = (int) Math.floor((sitesToSearch.size() / 3));
    Integer middle = (int) Math.floor((sitesToSearch.size() / 2));
    Integer end = (int) Math.floor((sitesToSearch.size() / 1));

    ArrayList<String> crawl_list1 = (ArrayList<String>)tmp.subList(0, start);
    ArrayList<String> crawl_list2 = (ArrayList<String>)tmp.subList(start+1, middle);
    ArrayList<String> crawl_list3 = (ArrayList<String>)tmp.subList(middle+1, end);

遗憾的是，这会引发以下错误:

Exception in thread "main" java.lang.ClassCastException: java.util.ArrayList$SubList cannot be cast to java.util.ArrayList

那么我怎样才能将它分成三个较小的ArrayList

tmp declaration:

public ArrayList<String> getExternalLinks(ArrayList<String> rootDomains){
ArrayList<String> result = new ArrayList<String>();

Document doc = null;
/*
 * Check if root is valid
 * Find search the site for internal links
 */
for (String root : rootDomains) {
    if (!(root == null) || !root.isEmpty() ||!root.contains("#")) {
        try {
            doc = Jsoup.connect(root)
                    .userAgent("Mozilla")
                    .get();
            result.addAll(findExternalLinks(findInternalLinks(doc,root),root)); 
        } catch (IOException e) {
            // TODO Auto-generated catch block
            e.printStackTrace();
        }
    }
}
System.out.println(result.size());
return result;

}

Answer 1

只需使用 List<String>作为 crawl_list1 的类型.

这只是"code against the interface"一般规则的应用.

确实没有充分的理由要求返回值 subList成为ArrayList (并且它没有意义，因为 subList 返回原始 ArrayList 的 View )。

如果您绝对需要ArrayList对象，那么您需要将内容复制到new ArrayList对象:

ArrayList<String> crawl_list1 = new ArrayList<String>(tmp.subList(0, start));
ArrayList<String> crawl_list2 = new ArrayList<String>(tmp.subList(start+1, middle));
ArrayList<String> crawl_list3 = new ArrayList<String>(tmp.subList(middle+1, end));