以LinkedList为键排序LinkedHashMap

Question

LinkedHashMap<LinkedList<Integer>, float[]> hm = new LinkedHashMap<>();
LinkedList<Integer> linkedlist;

linkedlist = new LinkedList<>();
linkedlist.add(10);
linkedlist.add(7);
hm.put(linkedlist, new float[]{0.14f, 1.2f, 85.01f});

linkedlist = new LinkedList<>();
linkedlist.add(0);
linkedlist.add(41);
hm.put(linkedlist, new float[]{10.3f, 50.05f, 9.9f});

linkedlist = new LinkedList<>();
linkedlist.add(210);
linkedlist.add(3);
hm.put(linkedlist, new float[]{17.0f, 4.0f, 2.1f});

现在我想要的输出是：

{0,41},{10.3f,50.05f,9.9f}

{10,7},{0.14f,1.2f,85.01f}

LinkedHashMap<LinkedList<Integer>, float[]> sortedMap = new LinkedHashMap<>();
hm.entrySet()
    .stream()
    .sorted(Map.Entry.comparingByKey())
    .forEachOrdered(x -> sortedMap.put(x.getKey(), x.getValue()));
hm = sortedMap;

for (int i = 0; i < hm.size(); i++) {
    if (i < 50) {
        for (Map.Entry<LinkedList<Integer>, float[]> entry : hm.entrySet()) {
            LinkedList<Integer> k = entry.getKey();
            System.out.println(k);
        }
    }
}

Answer 1

这种方法的问题是，您所拥有的键LinekdList 不是carable。您可以通过提供自定义的比较器来克服这一问题。

实现这种比较器的一种常见方法是比较匹配位置的整数，直到找到不相等的一对。如果其中一个列表比另一个短，并且在它结束之前没有击中其中的不匹配元素，则可以将其定义为比长列表“更小”：

Comparator<LinkedList<Integer>> cmp = new Comparator<LinkedList<Integer>>() {
    @Override
    public int compare(LinkedList<Integer> o1, LinkedList<Integer> o2) {
        Iterator<Integer> i1 = o1.iterator();
        Iterator<Integer> i2 = o2.iterator();

        // Iterate over the first list
        while (i1.hasNext()) {
            Integer x1 = i1.next();

            // If the second list has no more elements, it's "smaller"
            if (!i2.hasNext()) {
                return 1;
            }

            Integer x2 = i2.next();
            int cmp = Integer.compare(x1, x2);
            if (cmp != 0) {
                // If the matching items aren't equal, we know which list is "smaller"
                return cmp;
            }
        }

        // Done going over the first list.
        // If the second one still has items, it's "bigger"
        if (i2.hasNext()) {
            return -1;
        }

        // If not, both lists are equal
        return 0;
    }
};

一旦有了比较器，您只需使用它对映射的键进行排序：

LinkedHashMap<LinkedList<Integer>, float[]> sortedMap = new LinkedHashMap<>();
hm.entrySet()
        .stream()
        .sorted(Map.Entry.comparingByKey(cmp)) // Here
        .forEachOrdered(x -> sortedMap.put(x.getKey(), x.getValue()));
hm = sortedMap;