用于合并排序的链表的Python程序

合并排序通常是对链表进行排序的首选。链表的缓慢随机访问性能使得其他一些算法（如快速排序）表现不佳，而其他算法（如堆排序）则完全不可能。

设 head 为要排序的链表的第一个节点， headRef 为指向 head 的指针。请注意，我们需要在 MergeSort() 中引用 head，因为下面的实现会更改下一个链接以对链表（不是节点处的数据）进行排序，因此如果原始头部的数据不是链表中的最小值。

MergeSort(headRef)
1) If the head is NULL or there is only one element in the Linked List 
    then return.
2) Else divide the linked list into two halves.  
      FrontBackSplit(head, &a, &b); /* a and b are two halves */
3) Sort the two halves a and b.
      MergeSort(a);
      MergeSort(b);
4) Merge the sorted a and b (using SortedMerge() discussed here) 
   and update the head pointer using headRef.
     *headRef = SortedMerge(a, b);

Python3

# Python3 program to merge sort of linked list
  
# create Node using class Node.
class Node:
    def __init__(self, data):
        self.data = data
        self.next = None
  
class LinkedList:
    def __init__(self):
        self.head = None
      
    # push new value to linked list
    # using append method
    def append(self, new_value):
          
        # Allocate new node
        new_node = Node(new_value)
          
        # if head is None, initialize it to new node
        if self.head is None:
            self.head = new_node
            return
        curr_node = self.head
        while curr_node.next is not None:
            curr_node = curr_node.next
              
        # Append the new node at the end
        # of the linked list
        curr_node.next = new_node
          
    def sortedMerge(self, a, b):
        result = None
          
        # Base cases
        if a == None:
            return b
        if b == None:
            return a
              
        # pick either a or b and recur..
        if a.data <= b.data:
            result = a
            result.next = self.sortedMerge(a.next, b)
        else:
            result = b
            result.next = self.sortedMerge(a, b.next)
        return result
      
    def mergeSort(self, h):
          
        # Base case if head is None
        if h == None or h.next == None:
            return h
  
        # get the middle of the list 
        middle = self.getMiddle(h)
        nexttomiddle = middle.next
  
        # set the next of middle node to None
        middle.next = None
  
        # Apply mergeSort on left list 
        left = self.mergeSort(h)
          
        # Apply mergeSort on right list
        right = self.mergeSort(nexttomiddle)
  
        # Merge the left and right lists 
        sortedlist = self.sortedMerge(left, right)
        return sortedlist
      
    # Utility function to get the middle 
    # of the linked list 
    def getMiddle(self, head):
        if (head == None):
            return head
  
        slow = head
        fast = head
  
        while (fast.next != None and 
               fast.next.next != None):
            slow = slow.next
            fast = fast.next.next
              
        return slow
          
# Utility function to print the linked list 
def printList(head):
    if head is None:
        print(' ')
        return
    curr_node = head
    while curr_node:
        print(curr_node.data, end = " ")
        curr_node = curr_node.next
    print(' ')
      
# Driver Code
if __name__ == '__main__':
    li = LinkedList()
      
    # Let us create a unsorted linked list
    # to test the functions created. 
    # The list shall be a: 2->3->20->5->10->15 
    li.append(15)
    li.append(10)
    li.append(5)
    li.append(20)
    li.append(3)
    li.append(2)
      
    # Apply merge Sort 
    li.head = li.mergeSort(li.head)
    print ("Sorted Linked List is:")
    printList(li.head)
  
# This code is contributed by Vikas Chitturi

输出：

Sorted Linked List is: 
2 3 5 10 15 20

时间复杂度： O(n*log n)

空间复杂度： O(n*log n)

方法 2：这种方法更简单，使用 log n 空间。

合并排序（）：

1. 如果链表的大小为 1 则返回头部
2. 使用龟兔赛跑方法找到中间点
3. 将 mid 的下一个存储在 head2 中，即右子链表。
4. 现在使下一个中点为空。
5. 递归调用左右子链表的mergeSort()，存储左右链表的新头。
6. 给定左右子链表的新头参数，调用merge()，并存储合并后返回的最终头。
7. 返回合并链表的最终头。

合并（头1，头2）：

1. 取一个指针，说合并以将合并列表存储在其中并在其中存储一个虚拟节点。
2. 取一个指针 temp 并为其分配合并。
3. 如果 head1 的数据小于 head2 的数据，则将 head1 存储在 temp 的 next 中并将 head1 移动到 head1 的 next 中。
4. 否则将 head2 存储在 temp 的下一个并将 head2 移动到 head2 的下一个。
5. 将 temp 移到 temp 的下一个。
6. 重复步骤 3、4 和 5，直到 head1 不等于 null 并且 head2 不等于 null。
7. 现在将第一个或第二个链表的任何剩余节点添加到合并的链表中。
8. 返回合并后的下一个（将忽略虚拟并返回最终合并链表的头部）

Python3

# Python program for the above approach
  
  
# Node Class
class Node:
      
    def __init__(self,key):
        self.data=key
        self.next=None
  
# Function to merge sort
def mergeSort(head):
    if (head.next == None):
        return head
      
    mid = findMid(head)
    head2 = mid.next
    mid.next = None
    newHead1 = mergeSort(head)
    newHead2 = mergeSort(head2)
    finalHead = merge(newHead1, newHead2)
    return finalHead
  
# Function to merge two linked lists
def merge(head1,head2):
    merged = Node(-1)
      
    temp = merged
    # While head1 is not null and head2
    # is not null
    while (head1 != None and head2 != None):
        if (head1.data < head2.data):
            temp.next = head1
            head1 = head1.next
        else:
            temp.next = head2
            head2 = head2.next
        temp = temp.next
      
    # While head1 is not null
    while (head1 != None):
        temp.next = head1
        head1 = head1.next
        temp = temp.next
      
    # While head2 is not null
    while (head2 != None):
        temp.next = head2
        head2 = head2.next
        temp = temp.next
      
    return merged.next
  
# Find mid using The Tortoise and The Hare approach
def findMid(head):
    slow = head
    fast = head.next
    while (fast != None and fast.next != None):
        slow = slow.next
        fast = fast.next.next
    return slow
  
# Function to print list
def printList(head):
    while (head != None):
        print(head.data,end=" ")
        head=head.next
  
# Driver Code
head = Node(7)
temp = head
temp.next = Node(10);
temp = temp.next;
temp.next = Node(5);
temp = temp.next;
temp.next = Node(20);
temp = temp.next;
temp.next = Node(3);
temp = temp.next;
temp.next = Node(2);
temp = temp.next;
  
# Apply merge Sort
head = mergeSort(head);
print("
Sorted Linked List is: 
");
  
printList(head);
   
# This code is contributed by avanitrachhadiya2155

输出：

Sorted Linked List is: 
2 3 5 7 10 20 

时间复杂度：O(n*log n)

空间复杂度： O(log n)

有关详细信息，请参阅有关链接列表的合并排序的完整文章！