合并排序链接列表

时间:2008-08-11 11:43:27

标签: algorithm sorting linked-list mergesort

我最近在梳理一些基础知识,发现合并排序链表是一个相当不错的挑战。如果你有一个很好的实现,那么在这里展示它。

19 个答案:

答案 0 :(得分:78)

不知道为什么它应该是一个很大的挑战,因为它在这里说明,这是一个简单的Java实现,没有任何“聪明的技巧”。

//The main function
public static Node merge_sort(Node head) 
{
    if(head == null || head.next == null) 
        return head;

    Node middle = getMiddle(head);      //get the middle of the list
    Node left_head = head;
    Node right_head = middle.next; 
    middle.next = null;             //split the list into two halfs

    return merge(merge_sort(left_head), merge_sort(right_head));  //recurse on that
}

//Merge subroutine to merge two sorted lists
public static Node merge(Node a, Node b)
{
    Node dummyHead = new Node();

    for(Node current  = dummyHead; a != null && b != null; current = current.next;)
    {
        if(a.data <= b.data) 
        {
            current.next = a; 
            a = a.next; 
        }
        else
        { 
            current.next = b;
            b = b.next; 
        }

    }
    current.next = (a == null) ? b : a;
    return dummyHead.next;
}

//Finding the middle element of the list for splitting
public static Node getMiddle(Node head)
{
    if(head == null) 
        return head;

    Node slow = head, fast = head;

    while(fast.next != null && fast.next.next != null)
    {
        slow = slow.next;
        fast = fast.next.next;
    }
    return slow;
}

答案 1 :(得分:17)

更简单/更清晰的实现可能是递归实现,NLog(N)执行时间更清晰。

typedef struct _aList {
    struct _aList* next;
    struct _aList* prev; // Optional.
    // some data
} aList;

aList* merge_sort_list_recursive(aList *list,int (*compare)(aList *one,aList *two))
{
    // Trivial case.
    if (!list || !list->next)
        return list;

    aList *right = list,
          *temp  = list,
          *last  = list,
          *result = 0,
          *next   = 0,
          *tail   = 0;

    // Find halfway through the list (by running two pointers, one at twice the speed of the other).
    while (temp && temp->next)
    {
        last = right;
        right = right->next;
        temp = temp->next->next;
    }

    // Break the list in two. (prev pointers are broken here, but we fix later)
    last->next = 0;

    // Recurse on the two smaller lists:
    list = merge_sort_list_recursive(list, compare);
    right = merge_sort_list_recursive(right, compare);

    // Merge:
    while (list || right)
    {
        // Take from empty lists, or compare:
        if (!right) {
            next = list;
            list = list->next;
        } else if (!list) {
            next = right;
            right = right->next;
        } else if (compare(list, right) < 0) {
            next = list;
            list = list->next;
        } else {
            next = right;
            right = right->next;
        }
        if (!result) {
            result=next;
        } else {
            tail->next=next;
        }
        next->prev = tail;  // Optional.
        tail = next;
    }
    return result;
}

注意:这具有递归的Log(N)存储要求。绩效应与我发布的其他策略大致相当。通过在(list&amp;&amp; right)运行合并循环时可以实现潜在的优化,并且简单地附加剩余的列表(因为我们并不真正关心列表的结尾;知道它们的合并就足够了)

答案 2 :(得分:10)

严重依据来自http://www.chiark.greenend.org.uk/~sgtatham/algorithms/listsort.html

的优秀代码

略微修剪,整理:


typedef struct _aList {
    struct _aList* next;
    struct _aList* prev; // Optional.
       // some data
} aList;

aList *merge_sort_list(aList *list,int (*compare)(aList *one,aList *two))
{
    int listSize=1,numMerges,leftSize,rightSize;
    aList *tail,*left,*right,*next;
    if (!list || !list->next) return list;  // Trivial case

    do { // For each power of two<=list length
        numMerges=0,left=list;tail=list=0; // Start at the start

        while (left) { // Do this list_len/listSize times:
            numMerges++,right=left,leftSize=0,rightSize=listSize;
            // Cut list into two halves (but don't overrun)
            while (right && leftSize<listSize) leftSize++,right=right->next;
            // Run through the lists appending onto what we have so far.
            while (leftSize>0 || (rightSize>0 && right)) {
                // Left empty, take right OR Right empty, take left, OR compare.
                if (!leftSize)                  {next=right;right=right->next;rightSize--;}
                else if (!rightSize || !right)  {next=left;left=left->next;leftSize--;}
                else if (compare(left,right)<0) {next=left;left=left->next;leftSize--;}
                else                            {next=right;right=right->next;rightSize--;}
                // Update pointers to keep track of where we are:
                if (tail) tail->next=next;  else list=next;
                // Sort prev pointer
                next->prev=tail; // Optional.
                tail=next;          
            }
            // Right is now AFTER the list we just sorted, so start the next sort there.
            left=right;
        }
        // Terminate the list, double the list-sort size.
        tail->next=0,listSize<<=1;
    } while (numMerges>1); // If we only did one merge, then we just sorted the whole list.
    return list;
}

注意:这是O(NLog(N))保证,并使用O(1)资源(没有递归,没有堆栈,没有)。

答案 3 :(得分:6)

一种有趣的方法是维护一个堆栈,只有当堆栈上的列表具有相同数量的元素时才合并,否则推送列表,直到你用完传入列表中的元素,然后合并叠加。

答案 4 :(得分:2)

我一直在为优化这种算法的混乱而着迷,下面是我最终得出的结果。 Internet和StackOverflow上的很多其他代码都非常糟糕。有人试图获得列表的中间点,进行递归,为剩余节点设置多个循环,保持大量事物的数量 - 所有这些都是不必要的。 MergeSort自然适合链表,算法可以很漂亮和紧凑,但到达那个状态并不容易。

根据我所知,下面的代码保持最小数量的变量,并且具有算法所需的最小逻辑步骤数(即,不会使代码无法维护/不可读)。但是我没有尝试最小化LOC并保留尽可能多的空白以保持可读性。我通过相当严格的单元测试测试了这段代码。

请注意,此答案结合了其他答案https://stackoverflow.com/a/3032462/207661中的一些技巧。虽然代码在C#中,但转换为C ++,Java等应该是微不足道的。

SingleListNode<T> SortLinkedList<T>(SingleListNode<T> head) where T : IComparable<T>
{
    int blockSize = 1, blockCount;
    do
    {
        //Maintain two lists pointing to two blocks, left and right
        SingleListNode<T> left = head, right = head, tail = null;
        head = null; //Start a new list
        blockCount = 0;

        //Walk through entire list in blocks of size blockCount
        while (left != null)
        {
            blockCount++;

            //Advance right to start of next block, measure size of left list while doing so
            int leftSize = 0, rightSize = blockSize;
            for (;leftSize < blockSize && right != null; ++leftSize)
                right = right.Next;

            //Merge two list until their individual ends
            bool leftEmpty = leftSize == 0, rightEmpty = rightSize == 0 || right == null;
            while (!leftEmpty || !rightEmpty)
            {
                SingleListNode<T> smaller;
                //Using <= instead of < gives us sort stability
                if (rightEmpty || (!leftEmpty && left.Value.CompareTo(right.Value) <= 0))
                {
                    smaller = left; left = left.Next; --leftSize;
                    leftEmpty = leftSize == 0;
                }
                else
                {
                    smaller = right; right = right.Next; --rightSize;
                    rightEmpty = rightSize == 0 || right == null;
                }

                //Update new list
                if (tail != null)
                    tail.Next = smaller;
                else
                    head = smaller;
                tail = smaller;
            }

            //right now points to next block for left
            left = right;
        }

        //terminate new list, take care of case when input list is null
        if (tail != null)
            tail.Next = null;

        //Lg n iterations
        blockSize <<= 1;

    } while (blockCount > 1);

    return head;
}

兴趣点

  • 对于需要1等列表的空列表的情况,没有特殊处理。这些案例&#34;只是工作&#34;。
  • 很多&#34;标准&#34;算法文本有两个循环来遍历剩余元素,以处理一个列表比其他列表短的情况。上面的代码消除了它的需要。
  • 我们确保排序稳定
  • 作为热点的内部while循环平均每次迭代评估3个表达式,我认为这是最小的。

更新:@ ideasman42有translated above code to C/C++以及修改评论和更多改进的建议。上面的代码现在是最新的。

答案 5 :(得分:2)

我决定在这里测试一些例子,还有一个方法,最初是由Jonathan Cunningham在Pop-11中编写的。我用C#编写了所有方法,并与一系列不同的列表大小进行了比较。我比较了Raja R Harinath的Mono eglib方法,Shital Shah的C#代码,Jayadev的Java方法,David Gamble的递归和非递归版本,Ed Wynn的第一个C代码(这与我的样本数据集崩溃,我没有调试)和Cunningham的版本。完整代码:https://gist.github.com/314e572808f29adb0e41.git

Mono eglib基于与Cunningham类似的想法并具有相似的速度,除非列表恰好已经排序,在这种情况下Cunningham的方法要快得多(如果它的部分排序,eglib稍快一些)。 eglib代码使用固定表来保持合并排序递归,而Cunningham的方法通过使用递增的递归级别来工作 - 所以它开始不使用递归,然后是1-deep递归,然后是2-deep recursion,依此类推,排序需要多少步骤。我发现Cunningham代码更容易理解,并且没有猜测参与递归表有多大,所以它得到了我的投票。我在这个页面尝试的其他方法慢了两倍或更多。

这是Pop-11排序的C#端口:

/// <summary>
/// Sort a linked list in place. Returns the sorted list.
/// Originally by Jonathan Cunningham in Pop-11, May 1981.
/// Ported to C# by Jon Meyer.
/// </summary>
public class ListSorter<T> where T : IComparable<T> {
    SingleListNode<T> workNode = new SingleListNode<T>(default(T));
    SingleListNode<T> list;

    /// <summary>
    /// Sorts a linked list. Returns the sorted list.
    /// </summary>
    public SingleListNode<T> Sort(SingleListNode<T> head) {
        if (head == null) throw new NullReferenceException("head");
        list = head;

        var run = GetRun(); // get first run
        // As we progress, we increase the recursion depth. 
        var n = 1;
        while (list != null) {
            var run2 = GetSequence(n);
            run = Merge(run, run2);
            n++;
        }
        return run;
    }

    // Get the longest run of ordered elements from list.
    // The run is returned, and list is updated to point to the
    // first out-of-order element.
    SingleListNode<T> GetRun() {
        var run = list; // the return result is the original list
        var prevNode = list;
        var prevItem = list.Value;

        list = list.Next; // advance to the next item
        while (list != null) {
            var comp = prevItem.CompareTo(list.Value);
            if (comp > 0) {
                // reached end of sequence
                prevNode.Next = null;
                break;
            }
            prevItem = list.Value;
            prevNode = list;
            list = list.Next;
        }
        return run;
    }

    // Generates a sequence of Merge and GetRun() operations.
    // If n is 1, returns GetRun()
    // If n is 2, returns Merge(GetRun(), GetRun())
    // If n is 3, returns Merge(Merge(GetRun(), GetRun()),
    //                          Merge(GetRun(), GetRun()))
    // and so on.
    SingleListNode<T> GetSequence(int n) {
        if (n < 2) {
            return GetRun();
        } else {
            n--;
            var run1 = GetSequence(n);
            if (list == null) return run1;
            var run2 = GetSequence(n);
            return Merge(run1, run2);
        }
    }

    // Given two ordered lists this returns a list that is the
    // result of merging the two lists in-place (modifying the pairs
    // in list1 and list2).
    SingleListNode<T> Merge(SingleListNode<T> list1, SingleListNode<T> list2) {
        // we reuse a single work node to hold the result.
        // Simplifies the number of test cases in the code below.
        var prevNode = workNode;
        while (true) {
            if (list1.Value.CompareTo(list2.Value) <= 0) {
                // list1 goes first
                prevNode.Next = list1;
                prevNode = list1;
                if ((list1 = list1.Next) == null) {
                    // reached end of list1 - join list2 to prevNode
                    prevNode.Next = list2;
                    break;
                }
            } else {        // same but for list2
                prevNode.Next = list2;
                prevNode = list2;
                if ((list2 = list2.Next) == null) {
                    prevNode.Next = list1;
                    break;
                }
            }
        }

        // the result is in the back of the workNode
        return workNode.Next;
    }
}

答案 6 :(得分:2)

最简单的是来自 Gonnet + Baeza Yates Handbook of Algorithms。你用你想要的排序元素的数量来调用它,它递归地被一分为二,直到它达到一个大小一个列表的请求,然后你只需剥离原始列表的前面。这些都会合并到一个完整大小的排序列表中。

[请注意,第一篇文章中基于酷堆栈的文章称为在线合并,它在Knuth Vol 3的练习中得到了最少的提及]

答案 7 :(得分:2)

这是另一种递归版本。这不需要沿着列表进行拆分:我们提供一个指向head元素的指针(不是排序的一部分)和一个长度,递归函数返回一个指向排序列表末尾的指针。

element* mergesort(element *head,long lengtho)
{ 
  long count1=(lengtho/2), count2=(lengtho-count1);
  element *next1,*next2,*tail1,*tail2,*tail;
  if (lengtho<=1) return head->next;  /* Trivial case. */

  tail1 = mergesort(head,count1);
  tail2 = mergesort(tail1,count2);
  tail=head;
  next1 = head->next;
  next2 = tail1->next;
  tail1->next = tail2->next; /* in case this ends up as the tail */
  while (1) {
    if(cmp(next1,next2)<=0) {
      tail->next=next1; tail=next1;
      if(--count1==0) { tail->next=next2; return tail2; }
      next1=next1->next;
    } else {
      tail->next=next2; tail=next2;
      if(--count2==0) { tail->next=next1; return tail1; }
      next2=next2->next;
    }
  }
}

答案 8 :(得分:1)

链接列表的非递归合并排序的另一个示例,其中函数不是类的一部分。此示例代码和HP / Microsoft std :: list :: sort都使用相同的基本算法。自下而上的非递归合并排序,使用指向列表的第一个节点的小(26到32)指针数组,其中array [i]为0或指向大小为2的列表。在我的系统上,英特尔2600K 3.4ghz,它可以在大约1秒内对具有32位无符号整数的400万个节点进行排序。

NODE * MergeLists(NODE *, NODE *); /* prototype */

/* sort a list using array of pointers to list       */
/* aList[i] == NULL or ptr to list with 2^i nodes    */

#define NUMLISTS 32             /* number of lists */
NODE * SortList(NODE *pList)
{
NODE * aList[NUMLISTS];         /* array of lists */
NODE * pNode;
NODE * pNext;
int i;
    if(pList == NULL)           /* check for empty list */
        return NULL;
    for(i = 0; i < NUMLISTS; i++)   /* init array */
        aList[i] = NULL;
    pNode = pList;              /* merge nodes into array */
    while(pNode != NULL){
        pNext = pNode->next;
        pNode->next = NULL;
        for(i = 0; (i < NUMLISTS) && (aList[i] != NULL); i++){
            pNode = MergeLists(aList[i], pNode);
            aList[i] = NULL;
        }
        if(i == NUMLISTS)   /* don't go beyond end of array */
            i--;
        aList[i] = pNode;
        pNode = pNext;
    }
    pNode = NULL;           /* merge array into one list */
    for(i = 0; i < NUMLISTS; i++)
        pNode = MergeLists(aList[i], pNode);
    return pNode;
}

/* merge two already sorted lists                    */
/* compare uses pSrc2 < pSrc1 to follow the STL rule */
/*   of only using < and not <=                      */
NODE * MergeLists(NODE *pSrc1, NODE *pSrc2)
{
NODE *pDst = NULL;          /* destination head ptr */
NODE **ppDst = &pDst;       /* ptr to head or prev->next */
    if(pSrc1 == NULL)
        return pSrc2;
    if(pSrc2 == NULL)
        return pSrc1;
    while(1){
        if(pSrc2->data < pSrc1->data){  /* if src2 < src1 */
            *ppDst = pSrc2;
            pSrc2 = *(ppDst = &(pSrc2->next));
            if(pSrc2 == NULL){
                *ppDst = pSrc1;
                break;
            }
        } else {                        /* src1 <= src2 */
            *ppDst = pSrc1;
            pSrc1 = *(ppDst = &(pSrc1->next));
            if(pSrc1 == NULL){
                *ppDst = pSrc2;
                break;
            }
        }
    }
    return pDst;
}

答案 9 :(得分:1)

这是我对Knuth的“列表合并排序”的实现(来自TAOCP第3卷的算法5.2.4L,第2版)。我最后会添加一些评论,但这里有一个摘要:

在随机输入上,它比Simon Tatham的代码运行得快一点(参见Dave Gamble的非递归答案,带链接)但比Dave Gamble的递归代码慢一点。它比任何一个都难以理解。至少在我的实现中,它要求每个元素都有两个指向元素的指针。 (另一种选择是一个指针和一个布尔标志。)因此,它可能不是一个有用的方法。但是,一个与众不同的观点是,如果输入具有已经排序的长延伸,则运行速度非常快。

element *knuthsort(element *list)
{ /* This is my attempt at implementing Knuth's Algorithm 5.2.4L "List merge sort"
     from Vol.3 of TAOCP, 2nd ed. */
  element *p, *pnext, *q, *qnext, head1, head2, *s, *t;
  if(!list) return NULL;

L1: /* This is the clever L1 from exercise 12, p.167, solution p.647. */
  head1.next=list;
  t=&head2;
  for(p=list, pnext=p->next; pnext; p=pnext, pnext=p->next) {
    if( cmp(p,pnext) > 0 ) { t->next=NULL; t->spare=pnext; t=p; }
  }
  t->next=NULL; t->spare=NULL; p->spare=NULL;
  head2.next=head2.spare;

L2: /* begin a new pass: */
  t=&head2;
  q=t->next;
  if(!q) return head1.next;
  s=&head1;
  p=s->next;

L3: /* compare: */
  if( cmp(p,q) > 0 ) goto L6;
L4: /* add p onto the current end, s: */
  if(s->next) s->next=p; else s->spare=p;
  s=p;
  if(p->next) { p=p->next; goto L3; } 
  else p=p->spare;
L5: /* complete the sublist by adding q and all its successors: */
  s->next=q; s=t;
  for(qnext=q->next; qnext; q=qnext, qnext=q->next);
  t=q; q=q->spare;
  goto L8;
L6: /* add q onto the current end, s: */
  if(s->next) s->next=q; else s->spare=q;
  s=q;
  if(q->next) { q=q->next; goto L3; } 
  else q=q->spare;
L7: /* complete the sublist by adding p and all its successors: */
  s->next=p;
  s=t;
  for(pnext=p->next; pnext; p=pnext, pnext=p->next);
  t=p; p=p->spare;
L8: /* is this end of the pass? */
  if(q) goto L3;
  if(s->next) s->next=p; else s->spare=p;
  t->next=NULL; t->spare=NULL;
  goto L2;
}

答案 10 :(得分:1)

mono eglib中有一个非递归链表合并输入。

基本思想是各种合并的控制循环与二进制整数的按位递增相平行。 O(n)合并到“插入” n 节点到合并树中,并且这些合并的等级对应于递增的二进制数字。使用这个类比,只需要将合并树的 O(log n)节点实现为临时保持数组。

答案 11 :(得分:0)

我在这里看不到任何C ++解决方案。所以,就到这里。希望它可以帮助某人。

class Solution {
public:
    ListNode *merge(ListNode *left, ListNode *right){
        ListNode *head = NULL, *temp = NULL;
        // Find which one is the head node for the merged list
        if(left->val <= right->val){
            head = left, temp = left;
            left = left->next;
        }
        else{
            head = right, temp = right;
            right = right->next;
        }
        while(left && right){
            if(left->val <= right->val){
                temp->next = left;
                temp = left;
                left = left->next;
            }
            else{
                temp->next = right;
                temp = right;
                right = right->next;
            }
        }
        // If some elements still left in the left or the right list
        if(left)
            temp->next = left;
        if(right)
            temp->next = right;
        return head;
    }

    ListNode* sortList(ListNode* head){
        if(!head || !head->next)
            return head;

        // Find the length of the list
        int length = 0;
        ListNode *temp = head;
        while(temp){
            length++;
            temp = temp->next;
        }
        // Reset temp
        temp = head;
        // Store half of it in left and the other half in right
        // Create two lists and sort them
        ListNode *left = temp, *prev = NULL;
        int i = 0, mid = length / 2;
        // Left list
        while(i < mid){
            prev = temp;
            temp = temp->next;
            i++;
        }
        // The end of the left list should point to NULL
        if(prev)
            prev->next = NULL;
        // Right list
        ListNode  *right = temp;
        // Sort left list
        ListNode *sortedLeft = sortList(left);
        // Sort right list
        ListNode *sortedRight = sortList(right);
        // Merge them
        ListNode *sortedList = merge(sortedLeft, sortedRight);
        return sortedList;
    }
};

答案 12 :(得分:0)

这是链接列表上的合并排序的Java实现:

  
      
  • 时间复杂度:O(n.logn)
  •   
  • 空间复杂度:O(1)-链表上的合并排序实现避免了通常与O(n)相关的O(n)辅助存储成本   算法
  •   
class Solution
{
    public ListNode mergeSortList(ListNode head) 
    {
        if(head == null || head.next == null)
            return head;

        ListNode mid = getMid(head), second_head = mid.next; mid.next = null;

        return merge(mergeSortList(head), mergeSortList(second_head));
    }

    private ListNode merge(ListNode head1, ListNode head2)
    {
        ListNode result = new ListNode(0), current = result;

        while(head1 != null && head2 != null)
        {
            if(head1.val < head2.val)
            {
                current.next = head1;
                head1 = head1.next;
            }
            else
            {
                current.next = head2;
                head2 = head2.next;
            }
            current = current.next;
        }

        if(head1 != null) current.next = head1;
        if(head2 != null) current.next = head2;

        return result.next;
    }

    private ListNode getMid(ListNode head)
    {
        ListNode slow = head, fast = head.next;

        while(fast != null && fast.next != null)
        {
            slow = slow.next;
            fast = fast.next.next;
        }
        return slow;
    }
}

答案 13 :(得分:0)

嘿,我知道这是一个较晚的答案,但是得到了一个快速简单的答案。

该代码使用F#,但可以使用任何语言。由于这是ML系列的不常用语言,因此我会提出一些要点,以增强可读性。 F#是通过列表完成的嵌套。函数(嵌套部分)的最后一行代码是返回值。 (x,y)是一个元组,x :: xs是头x和尾xs的列表(其中xs可以为空),|>将最后一行的结果作为它右边的表达式的管道(增强可读性)和last(有趣的args->一些表达式)是lambda函数。

// split the list into a singleton list
let split list = List.map (fun x -> [x]) lst

// takes to list and merge them into a sorted list
let sort lst1 lst2 =
   // nested function to hide accumulator
   let rec s acc pair =
       match pair with
       // empty list case, return the sorted list
       | [], [] -> List.rev acc
       | xs, [] | [], xs ->
          // one empty list case, 
          // append the rest of xs onto acc and return the sorted list
          List.fold (fun ys y -> y :: ys) acc xs
          |> List.rev
       // general case
       | x::xs, y::ys ->
          match x < y with
          | true -> // cons x onto the accumulator
              s (x::acc) (xs,y::ys)
          | _ ->
              // cons y onto the accumulator
              s (y::acc) (x::xs,ys)

    s [] (lst1, lst2)  

let msort lst =
  let rec merge acc lst =
      match lst with
      | [] ->
          match acc with
          | [] -> [] // empty list case
          | _ -> merge [] acc
      | x :: [] -> // single list case (x is a list)
         match acc with
         | [] -> x // since acc are empty there are only x left, hence x are the sorted list.
         | _ -> merge [] (x::acc) // still need merging.
       | x1 :: x2 :: xs ->
           // merge the lists x1 and x2 and add them to the acummulator. recursiv call
           merge (sort x1 x2 :: acc) xs

   // return part
   split list // expand to singleton list list
   |> merge [] // merge and sort recursively.

重要的是要注意,这是完全尾递归的,因此没有堆栈溢出的可能性,并且通过一次将列表首先扩展到单例列表,我们降低了最坏开销上的常数。由于merge正在处理列表的列表,因此我们可以递归合并并排序内部列表,直到到达将所有内部列表排序到一个列表的修复点,然后我们将该列表返回,从而从列表列表折叠为列表再次。

答案 14 :(得分:0)

这是整段代码,展示了我们如何在java中创建链接列表并使用Merge排序对其进行排序。我在MergeNode类中创建节点,还有另一个MergesortLinklist类,其中有除法和合并逻辑。

class MergeNode {
    Object value;
    MergeNode next;

    MergeNode(Object val) {
        value = val;
        next = null;

    }

    MergeNode() {
        value = null;
        next = null;

    }

    public Object getValue() {
        return value;
    }

    public void setValue(Object value) {
        this.value = value;
    }

    public MergeNode getNext() {
        return next;
    }

    public void setNext(MergeNode next) {
        this.next = next;
    }

    @Override
    public String toString() {
        return "MergeNode [value=" + value + ", next=" + next + "]";
    }

}

public class MergesortLinkList {
    MergeNode head;
    static int totalnode;

    public MergeNode getHead() {
        return head;
    }

    public void setHead(MergeNode head) {
        this.head = head;
    }

    MergeNode add(int i) {
        // TODO Auto-generated method stub
        if (head == null) {
            head = new MergeNode(i);
            // System.out.println("head value is  "+head);
            return head;

        }
        MergeNode temp = head;

        while (temp.next != null) {
            temp = temp.next;
        }
        temp.next = new MergeNode(i);
        return head;

    }

    MergeNode mergesort(MergeNode nl1) {
        // TODO Auto-generated method stub

        if (nl1.next == null) {
            return nl1;
        }

        int counter = 0;

        MergeNode temp = nl1;

        while (temp != null) {
            counter++;
            temp = temp.next;

        }
        System.out.println("total nodes  " + counter);

        int middle = (counter - 1) / 2;

        temp = nl1;
        MergeNode left = nl1, right = nl1;
        int leftindex = 0, rightindex = 0;

        if (middle == leftindex) {
            right = left.next;
        }
        while (leftindex < middle) {

            leftindex++;
            left = left.next;
            right = left.next;
        }

        left.next = null;
        left = nl1;

        System.out.println(left.toString());
        System.out.println(right.toString());

        MergeNode p1 = mergesort(left);
        MergeNode p2 = mergesort(right);

        MergeNode node = merge(p1, p2);

        return node;

    }

    MergeNode merge(MergeNode p1, MergeNode p2) {
        // TODO Auto-generated method stub

        MergeNode L = p1;
        MergeNode R = p2;

        int Lcount = 0, Rcount = 0;

        MergeNode tempnode = null;

        while (L != null && R != null) {

            int val1 = (int) L.value;

            int val2 = (int) R.value;

            if (val1 > val2) {

                if (tempnode == null) {
                    tempnode = new MergeNode(val2);
                    R = R.next;
                } else {

                    MergeNode store = tempnode;

                    while (store.next != null) {
                        store = store.next;
                    }
                    store.next = new MergeNode(val2);

                    R = R.next;
                }

            } else {
                if (tempnode == null) {
                    tempnode = new MergeNode(val1);
                    L = L.next;
                } else {

                    MergeNode store = tempnode;

                    while (store.next != null) {
                        store = store.next;
                    }
                    store.next = new MergeNode(val1);

                    L = L.next;
                }

            }

        }

        MergeNode handle = tempnode;

        while (L != null) {

            while (handle.next != null) {

                handle = handle.next;

            }
            handle.next = L;

            L = null;

        }

        // Copy remaining elements of L[] if any
        while (R != null) {
            while (handle.next != null) {

                handle = handle.next;

            }
            handle.next = R;

            R = null;

        }

        System.out.println("----------------sorted value-----------");
        System.out.println(tempnode.toString());
        return tempnode;
    }

    public static void main(String[] args) {
        MergesortLinkList objsort = new MergesortLinkList();
        MergeNode n1 = objsort.add(9);
        MergeNode n2 = objsort.add(7);
        MergeNode n3 = objsort.add(6);
        MergeNode n4 = objsort.add(87);
        MergeNode n5 = objsort.add(16);
        MergeNode n6 = objsort.add(81);

        MergeNode n7 = objsort.add(21);
        MergeNode n8 = objsort.add(16);

        MergeNode n9 = objsort.add(99);
        MergeNode n10 = objsort.add(31);

        MergeNode val = objsort.mergesort(n1);

        System.out.println("===============sorted values=====================");
        while (val != null) {
            System.out.println(" value is  " + val.value);
            val = val.next;
        }
    }

}

答案 15 :(得分:0)

单个链表的经过测试的有效C++版本,基于投票最高的答案

单链接列表。h:

#pragma once
#include <stdexcept>
#include <iostream>
#include <initializer_list>
namespace ythlearn{
    template<typename T>
    class Linkedlist{
    public:
        class Node{
        public:
            Node* next;
            T elem;
        };
        Node head;
        int _size;
    public:
        Linkedlist(){
            head.next = nullptr;            
            _size = 0;
        }

        Linkedlist(std::initializer_list<T> init_list){
            head.next = nullptr;            
            _size = 0;
            for(auto s = init_list.begin(); s!=init_list.end(); s++){
                push_left(*s);
            }
        }

        int size(){
            return _size;
        }

        bool isEmpty(){
            return size() == 0;
        }

        bool isSorted(){
            Node* n_ptr = head.next;
            while(n_ptr->next != nullptr){
                if(n_ptr->elem > n_ptr->next->elem)
                    return false;
                n_ptr = n_ptr->next;
            }
            return true;
        }

        Linkedlist& push_left(T elem){
            Node* n = new Node;
            n->elem = elem;
            n->next = head.next;
            head.next = n;
            ++_size;
            return *this;
        }

        void print(){
                Node* loopPtr = head.next;
                while(loopPtr != nullptr){
                    std::cout << loopPtr->elem << " ";
                    loopPtr = loopPtr->next;
                }
                std::cout << std::endl;
        }

        void call_merge(){
            head.next = merge_sort(head.next);
        }

        Node* merge_sort(Node* n){
            if(n == nullptr || n->next == nullptr)
                return n;
            Node* middle = getMiddle(n);
            Node* left_head = n;
            Node* right_head = middle->next;
            middle->next = nullptr;
            return merge(merge_sort(left_head), merge_sort(right_head));
        }

        Node* getMiddle(Node* n){
            if(n == nullptr)
                return n;
            Node* slow, *fast;
            slow = fast = n;
            while(fast->next != nullptr && fast->next->next != nullptr){
                slow = slow->next;
                fast = fast->next->next;
            }
            return slow;
        }

        Node* merge(Node* a, Node* b){
            Node dummyHead;
            Node* current = &dummyHead;
            while(a != nullptr && b != nullptr){
                if(a->elem < b->elem){
                    current->next = a;
                    a = a->next;
                }else{
                    current->next = b;
                    b = b->next;
                }
                current = current->next;
            }
            current->next = (a == nullptr) ? b : a;
            return dummyHead.next;
        }

        Linkedlist(const Linkedlist&) = delete;
        Linkedlist& operator=(const Linkedlist&) const = delete;
        ~Linkedlist(){
            Node* node_to_delete;
            Node* ptr = head.next;
            while(ptr != nullptr){
                node_to_delete = ptr;
                ptr = ptr->next;
                delete node_to_delete;
            }

        }

    };
}

main.cpp:

#include <iostream>
#include <cassert>
#include "singlelinkedlist.h"
using namespace std;
using namespace ythlearn;

int main(){
    Linkedlist<int> l = {3,6,-5,222,495,-129,0};
    l.print();
    l.call_merge();
    l.print();
    assert(l.isSorted());
    return 0;
}

答案 16 :(得分:0)

最简单的Java实现:

  

时间复杂度:O(nLogn)n =节点数。链表的每次迭代都会使排序的较小链表的大小加倍。例如,在第一次迭代之后,链接列表将被分为两半。在第二次迭代之后,链接列表将被分为四个部分。它将继续按照链接列表的大小进行排序。这将使小型链表的大小增加O(logn)倍,从而达到原始链表的大小。之所以存在nlogn中的n,是因为链表的每次迭代所花费的时间与原始链表中节点的数量成正比。

class Node {
    int data;
    Node next;
    Node(int d) {
        data = d;
    }
}

class LinkedList {
    Node head;
    public Node mergesort(Node head) {
          if(head == null || head.next == null) return head;
          Node middle = middle(head), middle_next = middle.next;
          middle.next = null;
          Node left = mergesort(head), right = mergesort(middle_next), node = merge(left, right);
          return node;
    } 

    public Node merge(Node first, Node second) {
          Node node = null;
          if (first == null) return second;
          else if (second == null) return first;
          else if (first.data <= second.data) {
              node = first;
              node.next = merge(first.next, second);

          } else {
              node = second;
              node.next = merge(first, second.next);
          }
          return node;
    }

    public Node middle(Node head) {
          if (head == null) return head;
          Node second = head, first = head.next;
          while(first != null) {
              first = first.next;
              if (first != null) {
                 second = second.next;
                 first = first.next;
              }
          }
          return second;
    }

}

答案 17 :(得分:0)

这是使用 Swift编程语言的解决方案。

//Main MergeSort Function
func mergeSort(head: Node?) -> Node? {
   guard let head = head else { return nil }
   guard let _ = head.next else { return head }

   let middle = getMiddle(head: head)
   let left = head
   let right = middle.next

   middle.next = nil

   return merge(left: mergeSort(head: left), right: mergeSort(head: right))
}

//Merge Function
func merge(left: Node?, right: Node?) -> Node? {

   guard let left = left, let right = right else { return nil}

   let dummyHead: Node = Node(value: 0)

   var current: Node? = dummyHead
   var currentLeft: Node? = left
   var currentRight: Node? = right

   while currentLeft != nil && currentRight != nil {
       if currentLeft!.value < currentRight!.value {
        current?.next = currentLeft
        currentLeft = currentLeft!.next
       } else {
        current?.next = currentRight
        currentRight = currentRight!.next
       }
       current = current?.next
   }


   if currentLeft != nil {
        current?.next = currentLeft
   }

   if currentRight != nil {
        current?.next = currentRight
   }

   return dummyHead.next!
}

这是节点类 getMiddle方法

class Node { 
    //Node Class which takes Integers as value
    var value: Int
    var next: Node?
    
    init(value: Int) {
        self.value = value
    }
}

func getMiddle(head: Node) -> Node {
    guard let nextNode = head.next else { return head }
    
    var slow: Node = head
    var fast: Node? = head
    
    while fast?.next?.next != nil {
        slow = slow.next!
        fast = fast!.next?.next
    }
    
    
    return slow
}

答案 18 :(得分:-3)

public int[] msort(int[] a) {
    if (a.Length > 1) {
        int min = a.Length / 2;
        int max = min;

        int[] b = new int[min];
        int[] c = new int[max]; // dividing main array into two half arrays
        for (int i = 0; i < min; i++) {
            b[i] = a[i];
        }

        for (int i = min; i < min + max; i++) {
            c[i - min] = a[i];
        }

        b = msort(b);
        c = msort(c);

        int x = 0;
        int y = 0;
        int z = 0;

        while (b.Length != y && c.Length != z) {
            if (b[y] < c[z]) {
                a[x] = b[y];
                //r--
                x++;
                y++;
            } else {
                a[x] = c[z];
                x++;
                z++;
            }
        }

        while (b.Length != y) {
            a[x] = b[y];
            x++;
            y++;
        }

        while (c.Length != z) {
            a[x] = c[z];
            x++;
            z++;
        }
    }

    return a;
}