Java 如何将数组列表中的特定项移动到第一项
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How to move specific item in array list to the first item
提问by user782104
For example : A list
例如：一个列表
A B C D E
ABCDE
Given C , Switch to
给定 C ，切换到
C A B D E
CADE
Notice that the array size will change, some items may removed in run times
请注意，数组大小会发生变化，某些项目可能会在运行时删除
Collections.swap(url, url.indexOf(itemToMove), 0);
This statement is not working because it output C B A D E not C A B D E , how to fix it?
此语句不起作用，因为它输出 CBADE 而不是 CABDE ，如何解决？
Thanks.
谢谢。
采纳答案by Chris Hayes
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 Cdown by one.
你想要的是一个非常昂贵的ArrayList. 它需要将列表开头和Cdown位置之间的每个元素移动一个。
However, if you really want to do it:
但是，如果您真的想这样做：
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 Listimplementation 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.
如果这对您来说是一个频繁的操作，而随机访问的频率较低，您可以考虑切换到另一个List实现，例如LinkedList. 如果您非常关心元素的顺序，您还应该考虑列表是否是正确的数据结构。
回答by Venkata Krishna
Do this:
做这个：
- Removethe element from the list:
ArraylistObj.remove(object); - Addthe element backto the list at specific position:
ArrayListObj.add(position, Object);
- 从列表中删除元素：
ArraylistObj.remove(object); - 将元素添加回特定位置的列表：
ArrayListObj.add(position, Object);
As per your code use this :
根据您的代码使用此：
url.remove("C");
url.add(0,"C");
回答by hbsrud
The problem is, you swap C with A, so A B C D E becomes C B A D E.
问题是，你把 C 换成 A，所以 ABCDE 变成了 CBAD E。
You could try something like this:
你可以尝试这样的事情：
url.remove(itemToMove);
url.add(0, itemToMove);
Or if urlis a LinkedList:
或者如果url是LinkedList：
url.remove(itemToMove);
url.addFirst(itemToMove);
回答by sbc
This code will allow you to increase size of list, and insert elements without otherwise disturbing order of list
此代码将允许您增加列表的大小，并在不干扰列表顺序的情况下插入元素
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);
}
回答by Zulqarnain
Let say you have an array:
假设你有一个数组：
String[] arrayOne = new String[]{"A","B","C","D","E"};
Now you want to place the Cat index 0get the Cin another variable
现在，你要放置C在指数0得到C另一个变量
String characterC = arrayOne[2];
Now run the loop like following:
现在运行如下循环：
for (int i = (2 - 1); i >= 0; i--) {
arrayOne[i+1] = arrayOne[i];
}
Above 2is index of C. Now insert Cat index for example on 0
以上2是 的索引C。现在C在索引处插入例如0
arrayOne[0] = characterC;
Result of above loop will be like that:
上面循环的结果将是这样的：
arrayOne: {"C","A","B","D","E"}
The end, we achieve our goal.
最终，我们实现了目标。
回答by MYLS
Another solution, just keep swaping from 0to indexOf(itemToMove).
另一种解决方案，只需不断从0to交换即可indexOf(itemToMove)。
This is my Kotlin version:
这是我的 Kotlin 版本：
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.
抱歉，我不熟悉 Java，但学习了一点 Kotlin。但是算法是一样的。
