合并对链表进行排序

Question

我最近复习了一些基础知识，发现对链表进行合并排序是一个很好的挑战。如果你有一个很好的实现，那么在这里展示它。

Answer 1

不知道为什么它会是一个巨大的挑战，因为它在这里陈述，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; 
        }
        
    }
    dummyHead.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;
}

Answer 2

更简单/更清晰的实现可能是递归实现，它的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:这对递归有Log(N)存储要求。性能应该与我发布的其他策略大致相当。通过运行合并循环while (list && right)，并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；知道它们被合并就足够了)，这里有一个潜在的优化。

Answer 3

大量基于来自：http://www.chiark.greenend.org.uk/~sgtatham/algorithms/listsort.html的优秀代码

稍微修剪一下，整理一下：

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))保证的，并且使用O(1)资源(没有递归，没有堆栈，什么都没有)。