问题
I have an ArrayList
in Java
:
{"deleteItem", "createitem", "exportitem", "deleteItems", "createItems"}
I want to move all string which contains delete
to the end of the list, so I would get the next:
{"createitem", "exportitem", "createItems", "deleteItem", "deleteItems"}`
I can create two sublists - one for the words which contain the 'delete' word, and one for the others, and then merge them, but I search for a more efficient way.
回答1:
Use custom Comparator:
List<String> strings = Arrays.asList(
"deleteItem", "createitem", "exportitem", "deleteItems", "createItems"
);
Comparator<String> comparator = new Comparator<String>() {
@Override
public int compare(final String o1, final String o2) {
if (o1.contains("delete") && !o2.contains("delete")) {
return 1;
}else if (!o1.contains("delete") && o2.contains("delete")) {
return -1;
}
return 0;
}
};
Collections.sort(strings, comparator);
System.out.println(strings);
回答2:
If you want something efficient and need to remove elements in the beginning and middle of a List
I would suggest using a LinkedList
instead of a array list. That would avoid rewriting the underlying array for each remove operation.
Then, you simply iterate on the list, calling remove
and addLast
for any string that contains delete
.
Of course, this is only OK if there is nothing preventing you from replacing your ArrayList
with a LinkedList
.
回答3:
You only want to put the elements with delete at the end of the list so ordering is O(nlogn) while we could do this in one pass, in O(n) (although using a new list). We could create a new LinkedLIst and pass through the original list adding the elements with "delete" at the end and the others at the begginging.
LinkedList<String> orderedList = new LinkedList<>();
for(String e:originalList){
if (e.indexOf("delete")>=0) {
orderedList.addLast(e);
} else {
orderedList.addFirst(e);
}
}
This is faster than sorting.
来源:https://stackoverflow.com/questions/22905866/move-specific-items-to-the-end-of-a-list