Java将LinkedList的LinkedList按大小顺序排序

Question

I'm currently faced with this problem: I have a LinkedList that contains several LinkedLists containing longs, so:

LinkedList<LinkedList<Long>>() overalllList = new LinkedList<LinkedList<Long>();

After some code runs, the overall List gets filled with lists of longs of varying sizes. What I need to do is order the overallList so that it contains the lists of longs from smallest to largest.

I hope that makes sense.

So to clarify, I need this:

OverallList:
    LinkedList<Long> (size 2) - first
    LinkedList<Long> (size 245) - second
    LinkedList<Long> (size 1000) - third
    ...etc

I'm not sure if using Collections would do this, or if I need to look at a custom comparator. Any input or advice would be appreciated.

Thanks

Answer 1

Here is one example of a method to do that,

// A "size()" comparator 
private static Comparator<LinkedList<Long>> comp = new Comparator<LinkedList<Long>>() {
    @Override
    public int compare(LinkedList<Long> o1, LinkedList<Long> o2) {
        return new Integer((o1 == null) ? 0 : o1.size()).compareTo((o2 == null) ? 0 : o2.size());
    }
};
public static void main(String[] args) {
    // LinkedList<LinkedList<Long>>() overalllList = new LinkedList<LinkedList<Long>();
    // Note there is an extra () to the left of your overalllList.
    LinkedList<LinkedList<Long>> overalllList = new LinkedList<LinkedList<Long>>();
    LinkedList<Long> list3 = new LinkedList<Long>();
    LinkedList<Long> list2 = new LinkedList<Long>();
    LinkedList<Long> list1 = new LinkedList<Long>();

    for (long i = 0; i < 5; i++) { // 5, or 1000
        if (i < 2) {
            list1.add(i);
        }
        if (i < 3) { // 3, or 245.
            list2.add(i);
        }
        list3.add(i);
    }
    overalllList.add(list3);
    overalllList.add(list2);
    overalllList.add(list1);
    System.out.println("Before: " + overalllList);

    Collections.sort(overalllList, comp);
    System.out.println("After: " + overalllList);
}

Output is

Before: [[0, 1, 2, 3, 4], [0, 1, 2], [0, 1]]
After: [[0, 1], [0, 1, 2], [0, 1, 2, 3, 4]]

Answer 2

List<List<Long>> overalllList = new LinkedList<List<Long>>();
overalllList.add(Arrays.asList(1L, 2L, 3L));
overalllList.add(Arrays.asList(4L, 5L, 6L, 7L, 8L));
overalllList.add(Arrays.asList(9L));

Collections.sort(overalllList, new Comparator<List<Long>>() {
    @Override
    public int compare(List<Long> list1, List<Long> list2) {
        return list1.size() - list2.size();
    }
});