Java的Collections.sort（列表，比较器）的排序顺序是什么？从小到大还是从大到小？

Question

Apparently, it's not documented or I missed it.

Here 's the link to the documentation and below's the text as an image:

EDIT (17/5): I think too many confused this question to be a comparator question. It is not. The comparator compares between 2 elements. According to that comparison, the list sorted. How? Ascending or Descending?

I'll refine/simplify the question even further: If the comparator decides that element A is smaller than element B. In the sorted list , will element A be located at a lower index than element B?

Answer 1

The sort order is always ascending , where the Comparator defines which items are larger than others.

From the documentation for Collections.sort(List<T> list, Comparator<? super T> c) :

Sorts the specified list according to the order induced by the specified comparator.

From the documentation for Comparator.compare(T,T) :

Compares its two arguments for order. Returns a negative integer, zero, or a positive integer as the first argument is less than, equal to, or greater than the second.

Answer 2

You (or rather, your comparator) decides.

• If your Comparator 's compare(T o1, T o2) return a negative when o1 is less than o2 , you get ascending order ( demo on ideone ).
• If your Comparator 's compare(T o1, T o2) return a negative when o1 is greater than o2 , you get descending order ( demo on ideone ).

Another way of saying the same thing would be that sort assumes that the comparator orders the two items passed into it from smaller ( o1 ) to greater ( o2 ), and produces an ascending sort consistent with that ordering.

Answer 3

The documentation of Comparator.compareTo(o1, o2) method says

Compares its two arguments for order. Returns a negative integer, zero, or a positive integer as the first argument is less than, equal to, or greater than the second.

So if you want to sort from natural ordering , that is small to big, then you should write the implementation as defined in the documentation

public int compareTo(Integer o1, Integer o2) {
     int v1 = (o1);
     int v2 = (o2);
     if(v1 == v2) {
        return 0;
     }
     if(v1 < v2) {
        return -1; //return negative integer if first argument is less than second
     }
     return 1;
}

If you want the sorting to be in reverse order, that is big to small

public int compareTo(Integer o1, Integer o2) {
     int v1 = (o1);
     int v2 = (o2);
     if(v1 == v2) {
        return 0;
     }
     if(v1 < v2) {
        return 1;  //do the other way
     }
     return -1;
}

Answer 4

According to the documentation https://docs.oracle.com/javase/7/docs/api/java/util/Collections.html#sort(java.util.List,%20java.util.Comparator , the sort implementation for Collections.sort(list, comparator) is mergeSort.

Given the result produced by mergeSort is ascending ( https://en.wikipedia.org/wiki/Merge_sort ), the sort order of Collections.sort(list, comparator) is ascending.

That is to say if the comparator decides that element A is smaller than element B. In the sorted list, element A will be located at a smaller index than element B.