用Swift反转双链表

I have the code below to reverse a double linked list using Swift. However, I am confused on whether the swap function is swapping the currentNode with the node adjacent to it? or is it swapping it's two adjacent nodes?

我有下面的代码来反转使用Swift的双链表。但是，我对交换函数是否将currentNode与邻近的节点进行交换感到困惑？或者它是否交换它的两个相邻节点？

Example: Linked List values representation: 1 -> 2 -> 3

示例：链接列表值表示：1 - > 2 - > 3

Is it swapping 1 and 2 on the first run? or is it swapping 1 and 3? which values get swapped on the first run?

它是在第一次运行时交换1和2吗？还是交换1和3？哪些值在第一次运行时交换？

public func reverse() {
    var node = head
    while let currentNode = node {
        node = currentNode.next
        swap(&currentNode.next, &currentNode.previous)
        head = currentNode
    }
}

2 个解决方案

#1

You can add a print statement inside the loop (with a newline) to help you debug, or use the debugger to add a breakpoint.

您可以在循环内添加print语句（使用换行符）以帮助您进行调试，或使用调试器添加断点。

We can go through the function together to manually debug as well to improve understanding:

我们可以一起完成这个功能，手动调试以提高理解力：

First run:

第一次运行：

public func reverse() {
    var node = head
    while let currentNode = node {
        node = currentNode.next
        swap(&currentNode.next, &currentNode.previous)
        head = currentNode
    }
}

In this function:

在这个功能：

You assign node to head. They're both pointing to the node with the value 1
您将节点分配给head。它们都指向值为1的节点
if node exists (it does), then assign currentNode to node, so currentNode = node, therefore currentNode = 1 and node = 1 and head = 1 (all the same node)
如果节点存在（确实存在），则将currentNode分配给节点，因此currentNode = node，因此currentNode = 1且node = 1且head = 1（所有相同的节点）
(In the while): you say, node = currentNode.next. So now node is 2
（在while中）：你说，node = currentNode.next。所以现在节点是2
currentNode = 1 still. currentNode.previous is nil, and currentNode.next is 2.
currentNode = 1仍然。 currentNode.previous为nil，currentNode.next为2。
After the swap, the list looks like: 2 -> nil (->) 3 (I put the -> in parens because it doesn't actually "point" to 3 since it's nil).
在交换之后，列表看起来像：2 - > nil（ - >）3（我把 - >放在parens中，因为它实际上并不“指向”3，因为它是nil）。

Note that when I say = or is, above, like in node = 2, I mean "node variable is referring to the Node object with the value of 2"

请注意，当我说=或是，如上所述，就像在node = 2中一样，我的意思是“节点变量指的是值为2的Node对象”

So we're actually swapping the the first node's previous and next.

所以我们实际上交换了第一个节点的上一个节点和下一个节点。

#2

The swap call here swaps references to the next and previous element. Because when you're reverting a doubly linked list all "next" pointers should become "previous" and vice versa.

这里的交换调用交换对下一个和前一个元素的引用。因为当你恢复双向链表时，所有“下一个”指针应该变成“前一个”，反之亦然。

Take a look at this image for example

以此图片为例

And imagine that you need to change the marks on the arrows (next should become prev, prev should become next).

并且想象你需要改变箭头上的标记（下一个应该变为上一个，上一个应该成为下一个）。

#1