Python Program To Merge Two Sorted Lists (In-Place)

11 Jan, 2022
Given two sorted lists, merge them so as to produce a combined sorted list (without using extra space).Examples:
Input: head1: 5->7->9
        head2: 4->6->8 
Output: 4->5->6->7->8->9
Explanation: The output list is in sorted order.

Input: head1: 1->3->5->7
        head2: 2->4
Output: 1->2->3->4->5->7
Explanation: The output list is in sorted order.
There are different discussed different solutions in post below. Merge two sorted linked lists
Method 1 (Recursive):
Approach: The recursive solution can be formed, given the linked lists are sorted.
Compare the head of both linked lists.Find the smaller node among the two head nodes. The current element will be the smaller node among two head nodes.The rest elements of both lists will appear after that.Now run a recursive function with parameters, the next node of the smaller element, and the other head.The recursive function will return the next smaller element linked with rest of the sorted element. Now point the next of current element to that, i.e curr_ele->next=recursivefunction()Handle some corner cases. If both the heads are NULL return null.If one head is null return the other.
Compare the head of both linked lists.
Find the smaller node among the two head nodes. The current element will be the smaller node among two head nodes.
The rest elements of both lists will appear after that.
Now run a recursive function with parameters, the next node of the smaller element, and the other head.
The recursive function will return the next smaller element linked with rest of the sorted element. Now point the next of current element to that, i.e curr_ele->next=recursivefunction()
Handle some corner cases. If both the heads are NULL return null.If one head is null return the other.
If both the heads are NULL return null.
If one head is null return the other.
Python3
# Python3 program to merge two # sorted linked lists in-place.import mathclass Node:     def __init__(self, data):         self.data = data         self.next = None  # Function to create newNode in # a linkedlistdef newNode(key):    temp = Node(key)    temp.data = key    temp.next = None    return temp  # A utility function to print # linked listdef printList(node):    while (node != None):        print(node.data,               end = " ")        node = node.next  # Merges two given lists in-place. # This function mainly compares # head nodes and calls mergeUtil()def merge(h1, h2):    if (h1 == None):        return h2    if (h2 == None):        return h1      # start with the linked list    # whose head data is the least    if (h1.data < h2.data):        h1.next = merge(h1.next, h2)        return h1          else:        h2.next = merge(h1, h2.next)        return h2      # Driver Codeif __name__=='__main__':     head1 = newNode(1)    head1.next = newNode(3)    head1.next.next = newNode(5)      # 1.3.5 LinkedList created    head2 = newNode(0)    head2.next = newNode(2)    head2.next.next = newNode(4)      # 0.2.4 LinkedList created    mergedhead = merge(head1, head2)      printList(mergedhead)  # This code is contributed by Srathore
Output:  
0 1 2 3 4 5 
Complexity Analysis:
Time complexity:O(n). Only one traversal of the linked lists are needed.
Auxiliary Space:O(n). If the recursive stack space is taken into consideration.
Method 2 (Iterative):
Approach: This approach is very similar to the above recursive approach.
Traverse the list from start to end.If the head node of second list lies in between two nodes of the first list, insert it there and make the next node of second list the head. Continue this until there is no node left in both lists, i.e. both the lists are traversed.If the first list has reached end while traversing, point the next node to the head of second list.
Traverse the list from start to end.
If the head node of second list lies in between two nodes of the first list, insert it there and make the next node of second list the head. Continue this until there is no node left in both lists, i.e. both the lists are traversed.
If the first list has reached end while traversing, point the next node to the head of second list.
Note: Compare both the lists where the list with a smaller head value is the first list.
Python
# Python program to merge two # sorted linked lists in-place.# Linked List node class Node:     def __init__(self, data):         self.data = data         self.next = None  # Function to create newNode in # a linkedlistdef newNode(key):    temp = Node(0)    temp.data = key    temp.next = None    return temp  # A utility function to print # linked listdef printList(node):      while (node != None) :        print( node.data, end =" ")        node = node.next      # Merges two lists with headers as h1 and h2.# It assumes that h1's data is smaller than# or equal to h2's data.def mergeUtil(h1, h2):      # if only one node in first list    # simply point its head to second     # list    if (h1.next == None) :        h1.next = h2        return h1          # Initialize current and next     # pointers of both lists    curr1 = h1    next1 = h1.next    curr2 = h2    next2 = h2.next      while (next1 != None and            curr2 != None):               # if curr2 lies in between curr1         # and next1 then do curr1.curr2.next1        if ((curr2.data) >= (curr1.data) and            (curr2.data) <= (next1.data)):            next2 = curr2.next            curr1.next = curr2            curr2.next = next1              # now let curr1 and curr2 to point            # to their immediate next pointers            curr1 = curr2            curr2 = next2                  else:            # if more nodes in first list            if (next1.next) :                next1 = next1.next                curr1 = curr1.next                          # else point the last node of first list            # to the remaining nodes of second list            else:                next1.next = curr2                return h1      return h1  # Merges two given lists in-place. # This function mainly compares head # nodes and calls mergeUtil()def merge(h1, h2):      if (h1 == None):        return h2    if (h2 == None):        return h1      # Start with the linked list    # whose head data is the least    if (h1.data < h2.data):        return mergeUtil(h1, h2)    else:        return mergeUtil(h2, h1)  # Driver codehead1 = newNode(1)head1.next = newNode(3)head1.next.next = newNode(5)  # 1.3.5 LinkedList createdhead2 = newNode(0)head2.next = newNode(2)head2.next.next = newNode(4)  # 0.2.4 LinkedList createdmergedhead = merge(head1, head2)  printList(mergedhead)# This code is contributed by Arnab Kundu
Output:  
0 1 2 3 4 5 
Complexity Analysis:
Time complexity:O(n). As only one traversal of the linked lists is needed.
Auxiliary Space:O(1). As there is no space required.
Please refer complete article on Merge two sorted lists (in-place) for more details!
Accolite
Amazon
Belzabar
Brocade
FactSet
Flipkart
MakeMyTrip
Microsoft
OATS Systems
Oracle
Samsung
Synopsys
Linked List
Python Programs
Flipkart
Accolite
Amazon
Microsoft
Samsung
FactSet
MakeMyTrip
Oracle
Brocade
Synopsys
OATS Systems
Belzabar
Linked List
Writing code in comment?
Please use ide.geeksforgeeks.org,
generate link and share the link here.
Introduction to Data Structures
What is Data Structure: Types, Classifications and Applications
Types of Linked List
Circular Singly Linked List | Insertion
Find first node of loop in a linked list
Python program to convert a list to string
Defaultdict in Python
Python | Get dictionary keys as a list
Python | Convert a list to dictionary
Python | Convert string dictionary to dictionary