对整数数组列表的数组列表进行排序 [英] sort an arraylist of arraylist of integers

问题描述

我想对整数数组列表的数组列表进行排序，我需要帮助吗?

I am looking to sort an arraylist of arraylist of integers and I require help?

我被告知我需要实现比较器或可比较，然后使用 collection.sort 按顺序对列表列表进行排序...

I was informed that I need to implement comparator or comparable and then use the collection.sort to sort the list of list in order...

ArrayList<ArrayList<Integer>> g = new ArrayList<ArrayList<Integer>>()

If you look at the list of list as the following example:
C1 – 5,4,10
C2 – 3,2,1
C3 – 7,8,6
First it will be sorted like this:
C1 – 4,5,10
C2 – 1,2,3
C3 – 6,7,8
Then it will be sorted like this
C1 – 1,2,3
C2 – 4,5,6
C3 – 7,8,10

推荐答案

没有对空列表的错误检查，但在这里.

No error check for null lists, but here it is.

List<List<Integer>> list = Arrays.asList(Arrays.asList(10, 5, 4), 
        Arrays.asList(3, 2, 1), Arrays.asList(7, 8, 6));
for (List<Integer> l : list) {
    Collections.sort(l);
}
Collections.sort(list, new Comparator<List<Integer>>() {
    public int compare(List<Integer> o1, List<Integer> o2) {
        return o1.get(0).compareTo(o2.get(0));
    }
});
System.out.println(list);

Java 8 变得更加简洁:

With Java 8 it gets even more concise:

List<List<Integer>> list = Arrays.asList(Arrays.asList(10, 5, 4),
                Arrays.asList(3, 2, 1), Arrays.asList(7, 8, 6));
list.forEach(Collections::sort);
Collections.sort(list, (l1, l2) -> l1.get(0).compareTo(l2.get(0)));
System.out.println(list);

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