如何移动数组列表中的特定项目到第一个项目

Do this:

1. Remove the element from the list: ArraylistObj.remove(object);
2. Add the element back to the list at specific position: ArrayListObj.add(position, Object);

As per your code use this :

url.remove("C");
url.add(0,"C");

What you want is a very expensive operation in an ArrayList. It requires shifting every element between the beginning of the list and the location of C down by one.

However, if you really want to do it:

int index = url.indexOf(itemToMove);
url.remove(index);
url.add(0, itemToMove);

If this is a frequent operation for you, and random access is rather less frequent, you might consider switching to another List implementation such as LinkedList. You should also consider whether a list is the right data structure at all if you're so concerned about the order of elements.

The problem is, you swap C with A, so A B C D E becomes C B A D E.

You could try something like this:

url.remove(itemToMove);
url.add(0, itemToMove);

Or if url is a LinkedList:

url.remove(itemToMove);
url.addFirst(itemToMove);

This code will allow you to increase size of list, and insert elements without otherwise disturbing order of list

private void insert(double price){
for(int i = 0; i < keys.size(); i++){
if(price > keys.get(i)){
keys.add(null);
for(int j = keys.size()-1; j > i; j--){
Collections.swap(keys, j, j-1);
}
keys.add(price);
Collections.swap(keys, keys.size()-1, i);
keys.remove(keys.size()-1);
return;
}
}
keys.add(price);
}

Let say you have an array:

String[] arrayOne = new String[]{"A","B","C","D","E"};

Now you want to place the C at index 0 get the C in another variable

String characterC = arrayOne[2];

Now run the loop like following:

for (int i = (2 - 1); i >= 0; i--) {


arrayOne[i+1] = arrayOne[i];
}

Above 2 is index of C. Now insert C at index for example on 0

arrayOne[0] = characterC;

Result of above loop will be like that:

arrayOne: {"C","A","B","D","E"}

The end, we achieve our goal.

Another solution, just keep swaping from 0 to indexOf(itemToMove).

This is my Kotlin version:

val list = mutableListOf('A', 'B', 'C', 'D', 'E')
(0..list.indexOf('C')).forEach {
Collections.swap(list, 0, it)
}

Sorry I am unfamiliar with Java but learned a little Kotlin. But the algorithm is the same.

Kotlin ->

fun <T> MutableList<T>.move(item: T, newIndex: Int)  {
val currentIndex = indexOf(item)
if (currentIndex < 0) return
removeAt(currentIndex)
add(newIndex, item)
}

One more Kotlin solution based on other answers from this thread:

inline fun <T> List<T>.moveItemToFirstPosition(predicate: (T) -> Boolean): List<T> {
for (element in this.withIndex()) {
if (predicate(element.value)) {
return this.toMutableList().apply {
removeAt(element.index)
add(0, element.value)
}.toList()
}
}
return this
}

Usage:

var list = listOf("A", "B", "C", "D", "E")
list = list.moveItemToFirstPosition { it == "C" }

OR

inline fun <T> MutableList<T>.moveItemToFirstPosition(predicate: (T) -> Boolean) {
for (element in this.withIndex()) {
if (predicate(element.value)) {
removeAt(element.index)
add(0, element.value)
break
}
}
}

Usage:

val list = mutableListOf("A", "B", "C", "D", "E")
list.moveItemToFirstPosition { it == "C" }