我一直在寻找自然 mergesort实现（链接列表）一段时间，但没有运气。
这里我们有递归和迭代实现，但我不知道如何将其转换为自然的mergesort。
如何在最佳情况下检查运行以获得 O（n）复杂度？它不一定是C / C ++，可以是任何语言，甚至是Pseudocode。
谢谢。
答案 0 :(得分：0)
Wikipedia上有伪代码实现：
# 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
答案 1 :(得分：0)
这是我在F＃中的尝试。定期合并排序的实现以供参考：
// 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。在最好的情况下，这将是O（n），但最好的情况是输入列表按反向排序顺序。
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），其中最好的情况是现在输入列表已经排序。我否定了比较功能。排序列表将按相反的顺序构建，因此您需要在结束时将其反转。
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)
答案 2 :(得分：0)
我不确定什么是自然合并排序，但合并排序链表，我这样写：
[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;
}
答案 3 :(得分：-1)
我使用C＃
的原始算法实现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;
}
}