我在java中有一个双精度的列表，我想按降序排序数组列表。
输入数组列表如下:
List<Double> testList = new ArrayList();
testList.add(0.5);
testList.add(0.2);
testList.add(0.9);
testList.add(0.1);
testList.add(0.1);
testList.add(0.1);
testList.add(0.54);
testList.add(0.71);
testList.add(0.71);
testList.add(0.71);
testList.add(0.92);
testList.add(0.12);
testList.add(0.65);
testList.add(0.34);
testList.add(0.62);
输出应该是这样的
0.92
0.9
0.71
0.71
0.71
0.65
0.62
0.54
0.5
0.34
0.2
0.12
0.1
0.1
0.1
当前回答
你可以这样用
ArrayList<Group> groupList = new ArrayList<>();
Collections.sort(groupList, Collections.reverseOrder());
Collections.reverse(groupList);
其他回答
降:
Collections.sort(mArrayList, new Comparator<CustomData>() {
@Override
public int compare(CustomData lhs, CustomData rhs) {
// -1 - less than, 1 - greater than, 0 - equal, all inversed for descending
return lhs.customInt > rhs.customInt ? -1 : (lhs.customInt < rhs.customInt) ? 1 : 0;
}
});
|*|
import java.util.Collections;
|=>排序Asc顺序:
Collections.sort(NamAryVar);
|=>
Collections.sort(NamAryVar, Collections.reverseOrder());
|*|颠倒列表顺序:
Collections.reverse(NamAryVar);
yearList = arrayListOf()
for (year in 1950 until 2021) {
yearList.add(year)
}
yearList.reverse()
val list: ArrayList<String> = arrayListOf()
for (year in yearList) {
list.add(year.toString())
}
Collections.sort(testList);
Collections.reverse(testList);
这是你想要的。记住导入集合!
这里是集合的文档。
以下是一份简短的备忘单，涵盖了典型案例:
import static java.util.Comparator.comparing;
// sort
list.sort(naturalOrder());
// sort (reversed)
list.sort(reverseOrder());
// sort by field
list.sort(comparing(Type::getField));
// sort by field (reversed)
list.sort(comparing(Type::getField).reversed());
// sort by int field
list.sort(comparingInt(Type::getIntField));
// sort by double field (reversed)
list.sort(comparingDouble(Type::getDoubleField).reversed());
// sort by nullable field (nulls last)
list.sort(comparing(Type::getNullableField, nullsLast(naturalOrder())));
// two-level sort
list.sort(comparing(Type::getField1).thenComparing(Type::getField2));
