Java:如何按大小对多个ArrayList进行排序？

Question

我有9个不同ArrayList,我想有一个前5名的清单.

我正在考虑ArrayLists按尺寸对它们进行分类.

有可能吗？如果是这样,我怎么能实现呢？

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经过几次尝试后我终于开始工作,只想与大家分享.

最好是获得arraylist的大小并将其添加到大型arraylist

// creates an ArrayList that holds ArrayLists 
List allTheLists = new ArrayList(); 
allTheLists.add(pbaustraliaList.size());
allTheLists.add(pbotherList.size()); 
allTheLists.add(pbunitedStatesList.size()); 
allTheLists.add(pbunitedKingdomList.size()); 
allTheLists.add(pbchinaList.size()); 
allTheLists.add(pbgermanyList.size()); 
allTheLists.add(pbindiaList.size()); 
allTheLists.add(pbjapanList.size()); 
allTheLists.add(pbsingaporeList.size()); 
Comparator comparator = Collections.reverseOrder();
Collections.sort(allTheLists,comparator);

//display elements of ArrayList    
System.out.println("ArrayList elements after sorting in descending order : ");    
for(int i=0; i<allTheLists.size(); i++) {     
    System.out.println(allTheLists.get(i));   
}

Answer 1

你能做的是以下几点:

// this List of lists will need to contain 
// all of the ArrayLists you would like to sort
List<ArrayList> allTheLists; 
Collections.sort(allTheLists, new Comparator<ArrayList>(){
    public int compare(ArrayList a1, ArrayList a2) {
        return a2.size() - a1.size(); // assumes you want biggest to smallest
    }
});

这将按列表的长度对列表列表进行排序.排序列表中的第一个元素将是最长列表,最后一个元素将是最短列表.

然后,您可以遍历前5个列表以查看前5个列表.

一些链接供参考:

• 排序教程
• 收藏Javadoc
• 比较者Javadoc

根据您的ArrayLists存储方式,创建的代码List<ArrayList>看起来像这样:

// creates an ArrayList that holds ArrayLists
List<ArrayList> allTheLists = new ArrayList<ArrayList>();
allTheLists.add(yourList1);
allTheLists.add(yourList2);
...
allTheLists.add(yourList9);