Python Program For Inserting Node In The Middle Of The Linked List

10 Jul, 2022
Given a linked list containing n nodes. The problem is to insert a new node with data x at the middle of the list. If n is even, then insert the new node after the (n/2)th node, else insert the new node after the (n+1)/2th node.
Examples: 
Input : list: 1->2->4->5
        x = 3
Output : 1->2->3->4->5

Input : list: 5->10->4->32->16
        x = 41
Output : 5->10->4->41->32->16
Method 1(Using length of the linked list): Find the number of nodes or length of the linked using one traversal. Let it be len. Calculate c = (len/2), if len is even, else c = (len+1)/2, if len is odd. Traverse again the first c nodes and insert the new node after the cth node.  
Python3
# Python3 implementation to insert node# at the middle of a linked list # Node classclass Node:     # constructor to create a new node    def __init__(self, data):        self.data = data        self.next = None # function to insert node at the# middle of linked list given the headdef insertAtMid(head, x):     if(head == None): #if the list is empty        head = Node(x)    else:                 # create a new node for the value        # to be inserted        newNode = Node(x)         ptr = head        length = 0                 # calculate the length of the linked        # list        while(ptr != None):            ptr = ptr.next            length += 1         # 'count' the number of node after which        # the new node has to be inserted        if(length % 2 == 0):            count = length / 2        else:            (length + 1) / 2         ptr = head         # move ptr to the node after which        # the new node has to inserted        while(count > 1):            count -= 1            ptr = ptr.next         # insert the 'newNode' and adjust        # links accordingly        newNode.next = ptr.next        ptr.next = newNode # function to display the linked listdef display(head):    temp = head    while(temp != None):        print(str(temp.data), end = " ")        temp = temp.next # Driver Code # Creating the linked list 1.2.4.5head = Node(1)head.next = Node(2)head.next.next = Node(4)head.next.next.next = Node(5) print("Linked list before insertion: ", end = "")display(head) # inserting 3 in the middle of the linked list.x = 3insertAtMid(head, x) print("Linked list after insertion: " , end = "")display(head) # This code is contributed by Pranav Devarakonda
Output: 
Linked list before insertion: 1 2 4 5
Linked list after insertion: 1 2 3 4 5
Time Complexity: O(n)
Method 2(Using two pointers): Based on the tortoise and hare algorithm which uses two pointers, one known as slow and the other known as fast. This algorithm helps in finding the middle node of the linked list. It is explained in the front and black split procedure of this post. Now, you can insert the new node after the middle node obtained from the above process. This approach requires only a single traversal of the list. 
Python3
# Python implementation to insert node# at the middle of the linked list # Node Classclass Node :    def __init__(self, d):        self.data = d        self.next = None         class LinkedList:     # function to insert node at the    # middle of the linked list    def __init__(self):        self.head = None         # Function to insert a new node    # at the beginning    def push(self, new_data):        new_node = Node(new_data)        new_node.next = self.head        self.head = new_node             def insertAtMid(self, x):                 # if list is empty        if (self.head == None):            self.head = Node(x)         else:                         # get a new node            newNode = Node(x)             # assign values to the slow            # and fast pointers            slow = self.head            fast = self.head.next             while (fast != None and                   fast.next != None):                                 # move slow pointer to next node                slow = slow.next                 # move fast pointer two nodes                # at a time                fast = fast.next.next             # insert the 'newNode' and            # adjust the required links            newNode.next = slow.next            slow.next = newNode     # function to display the linked list    def display(self):        temp = self.head        while (temp != None):            print(temp.data, end = " "),            temp = temp.next # Driver Code # Creating the list 1.2.4.5ll = LinkedList()ll.push(5)ll.push(4)ll.push(2)ll.push(1)print("Linked list before insertion: "),ll.display() x = 3ll.insertAtMid(x) print("Linked list after insertion: "),ll.display() # This code is contributed by prerna saini
Output: 
Linked list before insertion: 1 2 4 5
Linked list after insertion: 1 2 3 4 5
Time Complexity: O(n)
Space complexity: O(n) where n is size of linked list
Please refer complete article on Insert node into the middle of the linked list for more details!
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