Python Program For Merge Sort Of Linked Lists
Merge sort is often preferred for sorting a linked list. The slow random-access performance of a linked list makes some other algorithms (such as quicksort) perform poorly, and others (such as heapsort) completely impossible.
Let the head be the first node of the linked list to be sorted and headRef be the pointer to head. Note that we need a reference to head in MergeSort() as the below implementation changes next links to sort the linked lists (not data at the nodes), so the head node has to be changed if the data at the original head is not the smallest value in the linked list.
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 |
Sorted Linked List is: 2 3 5 10 15 20
Time Complexity: O(n*log n)
Space Complexity: O(n*log n)
Approach 2: This approach is simpler and uses log n space.
mergeSort():
- If the size of the linked list is 1 then return the head
- Find mid using The Tortoise and The Hare Approach
- Store the next of mid in head2 i.e. the right sub-linked list.
- Now Make the next midpoint null.
- Recursively call mergeSort() on both left and right sub-linked list and store the new head of the left and right linked list.
- Call merge() given the arguments new heads of left and right sub-linked lists and store the final head returned after merging.
- Return the final head of the merged linkedlist.
merge(head1, head2):
- Take a pointer say merged to store the merged list in it and store a dummy node in it.
- Take a pointer temp and assign merge to it.
- If the data of head1 is less than the data of head2, then, store head1 in next of temp & move head1 to the next of head1.
- Else store head2 in next of temp & move head2 to the next of head2.
- Move temp to the next of temp.
- Repeat steps 3, 4 & 5 until head1 is not equal to null and head2 is not equal to null.
- Now add any remaining nodes of the first or the second linked list to the merged linked list.
- Return the next of merged(that will ignore the dummy and return the head of the final merged linked list)
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 |
Output:
Sorted Linked List is: 2 3 5 7 10 20
Time Complexity: O(n*log n)
Space Complexity: O(log n)
Please refer complete article on Merge Sort for Linked Lists for more details!
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