Python Program For Inserting Node In The Middle Of The Linked List
Last Updated :
19 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
class Node:
def __init__(self, data):
self.data = data
self.next = None
def insertAtMid(head, x):
if(head == None):
head = Node(x)
else:
newNode = Node(x)
ptr = head
length = 0
while(ptr != None):
ptr = ptr.next
length += 1
if(length % 2 == 0):
count = length / 2
else:
(length + 1) / 2
ptr = head
while(count > 1):
count -= 1
ptr = ptr.next
newNode.next = ptr.next
ptr.next = newNode
def display(head):
temp = head
while(temp != None):
print(str(temp.data), end = " ")
temp = temp.next
head = 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)
x = 3
insertAtMid(head, x)
print("
Linked list after insertion: " , end = "")
display(head)
|
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 the number of nodes in a linked list
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
class Node :
def __init__(self, d):
self.data = d
self.next = None
class LinkedList:
def __init__(self):
self.head = None
def push(self, new_data):
new_node = Node(new_data)
new_node.next = self.head
self.head = new_node
def insertAtMid(self, x):
if (self.head == None):
self.head = Node(x)
else:
newNode = Node(x)
slow = self.head
fast = self.head.next
while (fast != None and
fast.next != None):
slow = slow.next
fast = fast.next.next
newNode.next = slow.next
slow.next = newNode
def display(self):
temp = self.head
while (temp != None):
print(temp.data, end = " "),
temp = temp.next
ll = LinkedList()
ll.push(5)
ll.push(4)
ll.push(2)
ll.push(1)
print("Linked list before insertion: "),
ll.display()
x = 3
ll.insertAtMid(x)
print("
Linked list after insertion: "),
ll.display()
|
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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