自然归并排序一个链表

Question

我一直在寻找一个自然合并排序实现（链接列表），但没有成功。

Merge Sort a Linked List

这里我们同时拥有递归和迭代实现，但我不知道如何将其变成自然归并排序。

如何检查运行以在最佳情况下获得 O(n) 复杂度？它不必是 C/C++，可以是任何语言甚至是伪代码。

谢谢。

Answer 1

Wikipedia 上有一个伪代码实现：

 # Original data is on the input tape; the other tapes are blank
 function mergesort(input_tape, output_tape, scratch_tape_C, scratch_tape_D)
     while any records remain on the input_tape
         while any records remain on the input_tape
             merge( input_tape, output_tape, scratch_tape_C)
             merge( input_tape, output_tape, scratch_tape_D)
         while any records remain on C or D
             merge( scratch_tape_C, scratch_tape_D, output_tape)
             merge( scratch_tape_C, scratch_tape_D, input_tape)

 # take the next sorted chunk from the input tapes, and merge into the single given output_tape.
 # tapes are scanned linearly.
 # tape[next] gives the record currently under the read head of that tape.
 # tape[current] gives the record previously under the read head of that tape.
 # (Generally both tape[current] and tape[previous] are buffered in RAM ...)
 function merge(left[], right[], output_tape[])
     do
        if left[current] ≤ right[current]
            append left[current] to output_tape
            read next record from left tape
        else
            append right[current] to output_tape
            read next record from right tape
    while left[current] < left[next] and right[current] < right[next]
    if left[current] < left[next]
        append current_left_record to output_tape
    if right[current] < right[next]
        append current_right_record to output_tape
    return

Answer 2

这是我在 F# 中的尝试。一个正则归并排序的实现供参考：

// Sorts a list containing elements of type T.  Takes a comparison
// function comp that takes two elements of type T and returns -1
// if the first element is less than the second, 0 if they are equal,
// and 1 if the first element is greater than the second.
let rec sort comp = function 
| []  -> []  // An empty list is sorted
| [x] -> [x] // A single element list is sorted
| xs  ->
    // Split the list in half, sort both halves,
    // and merge the sorted halves.
    let half = (List.length xs) / 2
    let left, right = split half xs
    merge comp (sort comp left) (sort comp right)

现在尝试自然版本。最好的情况是 O(n)，但最好的情况是输入列表是反向排序的。

let rec sort' comp ls =

    // Define a helper function.  Streaks are stored in an accumulator.
    let rec helper accu = function
    | [] -> accu 
    | x::xs -> 
        match accu with
        // If we are not in a streak, start a new one
        | [] -> helper [x] xs 

        // If we are in a streak, check if x continues
        // the streak.
        | y::ys -> 
            if comp y x > 0 

            // x continues the streak so we add it to accu
            then helper (x::y::ys) xs

            // The streak is over. Merge the streak with the rest
            // of the list, which is sorted by calling our helper function on it.
            else merge comp accu (helper [x] xs)

    helper [] ls

第二次尝试。在最好的情况下，这也是 O(n)，现在最好的情况是输入列表已经排序。我否定了比较功能。排序后的列表将以相反的顺序构建，因此您需要在最后反转它。

let rec sort'' comp ls =
    // Flip the comparison function
    let comp' = fun x y -> -1 * (comp x y)
    let rec helper accu = function
    | [] -> accu
    | x::xs -> 
        match accu with
        | [] -> helper [x] xs
        | y::ys -> 
            if comp' y x > 0 
            then helper (x::y::ys) xs
            else merge comp' accu (helper [x] xs)

    // The list is in reverse sorted order so reverse it.
    List.rev (helper [] ls)

Answer 3

我不确定什么是自然归并排序，但是对于链表的归并排序，我是这样写的：

[Java 代码]

// Merge sort the linked list.
// From min to max.
// Time complexity = O(nlgn).
public static Node mergeSortLLFromMinToMax (Node head) {
    if (head == null || head.next == null) return head; // No need to sort.
    // Get the mid point of this linked list.
    Node prevSlower = head;
    Node slower = head;
    Node faster = head;
    while (faster != null && faster.next != null) {
        prevSlower = slower;
        slower = slower.next;
        faster = faster.next.next;
    }
    // Cut of the main linked list.
    prevSlower.next = null;

    // Do recursion.
    Node left = mergeSortLLFromMinToMax (head);
    Node right = mergeSortLLFromMinToMax (slower);

    // Merge the left and right part from min to max.
    Node currHead = new Node ();
    Node tempCurrHead = currHead;
    while (left != null && right != null) {
        if (left.data <= right.data) {
            // Add the elem of the left part into main linked list.
            tempCurrHead.next = left;
            left = left.next;
        } else {
            // Add the elem of the right part into main linked list.
            tempCurrHead.next = right;
            right = right.next;
        }
        tempCurrHead = tempCurrHead.next;
    }
    if (left != null) {
        // Add the remaining part of left part into main linked list.
        tempCurrHead.next = left;
        left = left.next;
        tempCurrHead = tempCurrHead.next;
    } else if (right != null) {
        // Add the remaining part of right part into main linked list.
        tempCurrHead.next = right;
        right = right.next;
        tempCurrHead = tempCurrHead.next;
    }

    return currHead.next;
}

Answer 4

我使用 C# 对算法的非常原始的实现

public static class LinkedListSort
{
    public static DataStructures.Linear.LinkedListNode<T> Sort<T>(DataStructures.Linear.LinkedListNode<T> firstNode) where T : IComparable<T>
    {
        if (firstNode == null)
            throw new ArgumentNullException();

        if (firstNode.Next == null)
            return firstNode;

        var head = firstNode;
        var leftNode = head;
        int iterNum = 0;

        while (leftNode != null)
        {
            //Let's start again from the begining
            leftNode = head;
            iterNum = 0;
            DataStructures.Linear.LinkedListNode<T> tailNode = null;

            while (leftNode != null)
            {
                //Let's get the left sublist

                //Let's find the node which devides sublist into two ordered sublists
                var sentinelNode = GetSentinelNode(leftNode);
                var rightNode = sentinelNode.Next;

                //If the right node is null it means that we don't have two sublist and the left sublist is ordered already
                //so we just add the rest sublist to the tail
                if (rightNode == null)
                {
                    if (tailNode == null)
                        break;
                    tailNode.Next = leftNode;
                    break;
                }

                sentinelNode.Next = null;

                //Let's find the node where the right sublist ends
                sentinelNode = GetSentinelNode(rightNode);
                var restNode = sentinelNode.Next;
                sentinelNode.Next = null;

                DataStructures.Linear.LinkedListNode<T> newTailNode = null;

                //Merging of two ordered sublists   
                var mergedList = Merge(leftNode, rightNode, ref newTailNode);
                //If we're at the beginning of the list the head of the merged sublist becomes the head of the list
                if (iterNum == 0)                   
                    head = mergedList;                  
                else //add the                  
                    tailNode.Next = mergedList;                     

                tailNode = newTailNode;
                leftNode = restNode;
                iterNum++;
            }
            if (iterNum == 0)
                break;
        }
        return head;
    }

    /// <summary>
    /// Merges two ordered sublists   
    /// </summary>
    /// <typeparam name="T"></typeparam>
    /// <param name="aNode">Left part of sublist</param>
    /// <param name="bNode">Right part of sublist</param>
    /// <param name="tailNode">Tail node of the merged list</param>
    /// <returns>The result of merging</returns>
    private static DataStructures.Linear.LinkedListNode<T> Merge<T>(DataStructures.Linear.LinkedListNode<T> leftNode,                                                                       
                                                                    DataStructures.Linear.LinkedListNode<T> rightNode,
                                                                    ref DataStructures.Linear.LinkedListNode<T> tailNode) where T : IComparable<T>
    {
        var dummyHead = new DataStructures.Linear.LinkedListNode<T>();
        var curNode = dummyHead;

        while (leftNode != null || rightNode != null)
        {
            if (rightNode == null)
            {
                curNode.Next = leftNode;
                leftNode = leftNode.Next;
            }
            else if (leftNode == null)
            {
                curNode.Next = rightNode;
                rightNode = rightNode.Next;
            }
            else if (leftNode.Value.CompareTo(rightNode.Value) <= 0)
            {
                curNode.Next = leftNode;
                leftNode = leftNode.Next;
            }
            else
            {
                curNode.Next = rightNode;
                rightNode = rightNode.Next;
            }
            curNode = curNode.Next;
        }
        tailNode = curNode;
        return dummyHead.Next;
    }

    /// <summary>
    /// Returns the sentinel node
    /// </summary>
    /// <typeparam name="T"></typeparam>
    /// <param name="firstNode"></param>
    /// <returns></returns>
    private static DataStructures.Linear.LinkedListNode<T> GetSentinelNode<T>(DataStructures.Linear.LinkedListNode<T> firstNode) where T : IComparable<T>
    {
        var curNode = firstNode;

        while (curNode != null && curNode.Next != null && curNode.Value.CompareTo(curNode.Next.Value) <= 0)             
            curNode = curNode.Next;

        return curNode;
    }
}