排序整数的arraylist的arraylist

我想整理一个整数的arraylist的arraylist，我需要帮助？

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

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

Answer 1

没有错误检查空列表，但在这里。

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

使用Java 8，它变得更加简洁：

 List> 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); 

Answer 2

您可以单独对每个列表进行排序。 Collections.sort(collection)将自动按升序对整数进行排序。

Answer 3

如果排序没有你需要的，你可以试试这个算法：

 package drawFramePackage; import java.awt.geom.AffineTransform; import java.util.ArrayList; import java.util.ListIterator; import java.util.Random; public class QuicksortAlgorithm { ArrayList affs; ListIterator li; Integer count, count2; /** * @param args */ public static void main(String[] args) { new QuicksortAlgorithm(); } public QuicksortAlgorithm(){ count = new Integer(0); count2 = new Integer(1); affs = new ArrayList(); for (int i = 0; i <= 128; i++){ affs.add(new AffineTransform(1, 0, 0, 1, new Random().nextInt(1024), 0)); } affs = arrangeNumbers(affs); printNumbers(); } public ArrayList arrangeNumbers(ArrayList list){ while (list.size() > 1 && count != list.size() - 1){ if (list.get(count2).getTranslateX() > list.get(count).getTranslateX()){ list.add(count, list.get(count2)); list.remove(count2 + 1); } if (count2 == list.size() - 1){ count++; count2 = count + 1; } else{ count2++; } } return list; } public void printNumbers(){ li = affs.listIterator(); while (li.hasNext()){ System.out.println(li.next()); } } }