c# - 如何从合并排序中获得O(n log(n))？

Question

如何知道我在链表上的合并排序实现中是否使用 O(nlog(n)) 。我应该向 O(nlog(n)) 输入什么才能知道我实现合并排序的时间。

public static LinkedListNode<T> MergeSortLL<T>(LinkedListNode<T> Head) where T : IComparable<T>
{

    if (Head?.Next == null)
    {
        return Head;
    }
    var middle = GetMiddle(Head);
    var half = middle.Next;
    middle.Next = null;

    Head = Merge(MergeSortLL(Head), MergeSortLL(half));
    return Head;
}

public static LinkedListNode<T> Merge<T>(LinkedListNode<T> Left, LinkedListNode<T> Right) where T : IComparable<T>
{

    var mHead = new LinkedListNode<T>(default(T));
    LinkedListNode<T> curr = mHead;

    while (Left != null && Right != null)
    {
        if (Left.Value.CompareTo(Right.Value) <= 0)
        {
            curr.Next = Left;
            Left = Left.Next;
        }
        else
        {
            curr.Next = Right;
            Right = Right.Next;
        }
        curr = curr.Next;
    }
    curr.Next = (Left == null) ? Right : Left;

    return mHead.Next;
}

public static LinkedListNode<T> GetMiddle<T>(LinkedListNode<T> Head) where T : IComparable<T>
{
    if (Head == null)
    {
        return Head;
    }

    LinkedListNode<T> slow, fast;
    slow = fast = Head;

    while (fast.Next?.Next != null)
    {
        slow = slow.Next;
        fast = fast.Next.Next;
    }
    return slow;
}

Answer 1

请注意，已知该算法执行 O(N*log(N)) 比较。如果您想确认这一点(因为它已被证明)，您可以放置​​一个计数器并在每次比较时递增它。

例如

public static int Comparisons;

public static LinkedListNode<T> Merge<T>(LinkedListNode<T> Left, LinkedListNode<T> Right) where T : IComparable<T>
{
    // ...
    Comparisons++;
    if (Left.Value.CompareTo(Right.Value) <= 0)
    // ...
}

并检查

LinkedListNode<int> head = ...;
Comparisons = 0;
head = MergeSortLL(head);
Debug.Print(Comparisons);

但请注意，该算法也具有显着的隐藏成本(即使使用慢/快指针技术，也要找到中间值)。