[英]Java Generic Doubly Linked List Swap
http://users.cis.fiu.edu/~weiss/dsaajava3/code/MyLinkedList.java http://users.cis.fiu.edu/~weiss/dsaajava3/code/MyLinkedList.java
Above link shows the code that I want to modify in order to have swap method from the doubly linked list. 上面的链接显示了我想要修改的代码,以便具有双向链接列表中的swap方法。 I apologize in advance using URL in place due being 300 line of code pasted here is not very readable.
我预先使用URL表示歉意,因为此处粘贴的300行代码不太可读。
For the output, I'd like to just have a single list output, for example : 对于输出,我只想有一个列表输出,例如 :
[ 28 27 **21** 25 24 23 22 **26** 20 0 1 2 3 4 5 6 7 8 ] //(2, 7) swap
Below is my attempt (part of the code to include/modify to original) : 以下是我的尝试(部分代码包含/修改为原始代码) :
swap method: 交换方法:
public AnyType swap( int idx, int idx2)
{
return swap( getNode( idx ), getNode ( idx2 ) );
}
private AnyType swap( Node<AnyType> p, Node<AnyType> p2)
{
Node<AnyType> temp = p; //realize it doesn't work due to linking
p = p2;
p2 = temp;
return p.data;
}
main: 主要:
class TestLinkedList
{
public static void main( String [ ] args )
{
MyLinkedList<Integer> lst = new MyLinkedList<>( );
for( int i = 0; i < 10; i++ )
lst.add( i );
for( int i = 20; i < 30; i++ )
lst.add( 0, i );
lst.remove( 0 );
lst.remove( lst.size( ) - 1 );
lst.swap(2, 7);
System.out.println("swap: " + lst);
}
}
output 输出
swap: [ 28 27 26 25 24 23 22 21 20 0 1 2 3 4 5 6 7 8 ]
21
I realize that my approach to solution is not the use of linked list. 我意识到解决问题的方法不是使用链表。 I've been analyzing the code that I am to modify;
我一直在分析要修改的代码; however am having hard time with the pointers and the doubly linked list to understand the coding part (not the concept).
但是,在使用指针和双向链表时很难理解编码部分(不是概念)。 Please explain how I should approach to solution.
请解释我应该如何解决。 Thanks.
谢谢。
EDIT 编辑
/* Swaps idx and idx2 nodes
* @param idx the index of the object
* @param idx2 the index of the object
*/
public void swap( int idx, int idx2)
{
swap( getNode( idx ), getNode ( idx2 ) );
if( idx < 0 || idx >= size( ) || idx2 < 0 || idx2 >= size( )){
throw new IndexOutOfBoundsException( "swap first index: " + idx + ", second index: " + idx2 );
}
}
/**
* Swaps node and data at p and p2
* @param p the index of the object
* @param p2 the index of the object
*/
public void swap( Node<AnyType> p, Node<AnyType> p2)
{
Node<AnyType> temp = p;
p = p2;
p2 = temp;
temp = p.next.prev;
p.next.prev = p2.next.prev;
p2.next.prev = temp;
AnyType dataTemp = p.data;
p.data = p2. data;
p2.data = dataTemp;
}
output 输出
[ 29 28 27 26 25 24 23 22 21 20 0 1 2 3 4 5 6 7 8 9 ]
swap: [ 29 28 22 26 25 24 23 27 21 20 0 1 2 3 4 5 6 7 8 9 ]
It is now working. 现在正在工作。 Did I do it right?
我做对了吗? However I get compile error with.
但是我得到编译错误。 I assume it has to do with
public void swap( Node<AnyType> p, Node<AnyType> p2)
method which has warning: exporting non-public type through public API
. 我假设它与
public void swap( Node<AnyType> p, Node<AnyType> p2)
方法有关,该方法具有warning: exporting non-public type through public API
。 What can I do here to fix it? 在这里我该怎么办?
Let's say your input is: 假设您的输入是:
[ 1 2 3 4 5 6 7 8 9 ]
and you want to swap 3 and 7: 而您想交换3和7:
[ 1 2 >7< 4 5 6 >3< 8 9 ]
This means you need to update all these values: 这意味着您需要更新所有这些值:
2.next = 7
7.prev = 2
7.next = 4
4.prev = 7
6.next = 3
3.prev = 6
3.next = 8
8.prev = 3
That is a total of 8 assignments, not counting assignment to temporary variables. 总共有8个分配,不包括对临时变量的分配。
What if you wanted to swap 5 and 6 instead: 如果您想交换5和6怎么办:
[ 1 2 3 4 >6< >5< 7 8 9 ]
4.next = 6
6.prev = 4
6.next = 5
5.prev = 6
5.next = 7
7.prev = 5
Hmmm... That was only 6 assignments. 嗯...那只是6个作业。
What if you wanted to swap first and last element: 如果要交换第一个和最后一个元素怎么办:
[ >9< 2 3 4 5 6 7 8 >1< ]
You didn't share this part, but I'll assume you have both a head and a tail reference, so: 您没有共享这部分,但是我假设您同时具有头和尾引用,所以:
head = 9
9.prev = null
9.next = 2
2.prev = 9
8.next = 1
1.prev = 8
1.next = null
tail = 1
Whoa, that was different. 哇,那不一样。
Now you just need to code all that, taking care to handle all combinations. 现在,您只需要编写所有代码,小心处理所有组合。
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