Python program to find middle of a linked list using one traversal
Last Updated :
23 Apr, 2023
Given a singly linked list, find the middle of the linked list. Given a singly linked list, find the middle of the linked list. For example, if the given linked list is 1->2->3->4->5 then the output should be 3.
Method 1: Traverse the whole linked list and count the no. of nodes. Now traverse the list again till count/2 and return the node at count/2.
Python3
class Node:
def __init__(self, data):
self.data = data
self.next = None
class NodeOperation:
def pushNode(self, head_ref, data_val):
new_node = Node(data_val)
new_node.next = head_ref
head_ref = new_node
return head_ref
def printNode(self, head):
while (head != None):
print('%d->' % head.data, end="")
head = head.next
print("NULL")
def getLen(self, head):
temp = head
len = 0
while (temp != None):
len += 1
temp = temp.next
return len
def printMiddle(self, head):
if head != None:
len = self.getLen(head)
temp = head
midIdx = len // 2
while midIdx != 0:
temp = temp.next
midIdx -= 1
print('The middle element is: ', temp.data)
head = None
temp = NodeOperation()
head = temp.pushNode(head, 5)
head = temp.pushNode(head, 4)
head = temp.pushNode(head, 3)
head = temp.pushNode(head, 2)
head = temp.pushNode(head, 1)
temp.printNode(head)
temp.printMiddle(head)
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Output
1->2->3->4->5->NULL
The middle element is: 3
Time Complexity: O(n) where n is no of nodes in linked list
Auxiliary Space: O(1)
Method 2: Traverse linked list using two pointers. Move one pointer by one and another pointer by two. When the fast pointer reaches the end slow pointer will reach middle of the linked list.
Implementation:
Python3
class Node:
def __init__(self, data):
self.data = data
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 printMiddle(self):
slow_ptr = self.head
fast_ptr = self.head
if self.head is not None:
while (fast_ptr is not None and fast_ptr.next is not None):
fast_ptr = fast_ptr.next.next
slow_ptr = slow_ptr.next
print("The middle element is: ", slow_ptr.data)
list1 = LinkedList()
list1.push(5)
list1.push(4)
list1.push(2)
list1.push(3)
list1.push(1)
list1.printMiddle()
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Output
The middle element is: 2
Time Complexity: O(N) where N is the number of nodes in Linked List.
Auxiliary Space: O(1)
Method 3: Initialized the temp variable as head Initialized count to Zero Take loop till head will become Null(i.e end of the list) and increment the temp node when count is odd only, in this way temp will traverse till mid element and head will traverse all linked list. Print the data of temp.
Implementation:
Python3
class Node:
def __init__(self, value):
self.data = value
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 printMiddle(self):
temp = self.head
count = 0
while self.head:
if (count & 1):
temp = temp.next
self.head = self.head.next
count += 1
print(temp.data)
llist = LinkedList()
llist.push(1)
llist.push(20)
llist.push(100)
llist.push(15)
llist.push(35)
llist.printMiddle()
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Complexity Analysis:
- Time complexity: O(N) where N is the size of the given linked list
- Auxiliary Space: O(1) because it is using constant space
METHOD 4:Using a list to store elements
APPROACH:
The above Python code implements a program that finds the middle element of a linked list using a single traversal. It defines two classes: Node and LinkedList. Node class has two instance variables: data and next. LinkedList class has an instance variable head which points to the first node of the linked list.
ALGORITHM:
1.Initialize a count variable to 0 and two pointers, middle_node, and current_node to the head of the linked list.
2.Traverse the linked list using current_node and increment the count variable by 1 in each iteration.
3.If the count variable is odd, move the middle_node pointer to the next node.
4.Return the data of the middle_node pointer as it points to the middle element.
Python3
class Node:
def __init__(self, data):
self.data = data
self.next = None
class LinkedList:
def __init__(self):
self.head = None
def insert(self, data):
new_node = Node(data)
if self.head is None:
self.head = new_node
else:
current_node = self.head
while current_node.next is not None:
current_node = current_node.next
current_node.next = new_node
def print_list(self):
current_node = self.head
while current_node is not None:
print(current_node.data, end="->")
current_node = current_node.next
print("NULL")
def find_middle(self):
elements = []
current_node = self.head
while current_node is not None:
elements.append(current_node.data)
current_node = current_node.next
middle_index = len(elements) // 2
return elements[middle_index]
linked_list = LinkedList()
linked_list.insert(1)
linked_list.insert(2)
linked_list.insert(3)
linked_list.insert(4)
linked_list.insert(5)
linked_list.print_list()
middle = linked_list.find_middle()
print("The middle element is:", middle)
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Output
1->2->3->4->5->NULL
The middle element is: 3
Time complexity: O(n), where n is the number of nodes in the linked list. We traverse the linked list only once.
Space complexity: O(1). We use a constant amount of extra space to store pointers and variables.
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