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?
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.
You (or rather, your comparator) decides.
Comparator
's compare(T o1, T o2)
return a negative when o1
is less than o2
, you get ascending order ( demo on ideone ). 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.
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;
}
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.
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