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.
If there are even nodes, then there would be two middle nodes, we need to print the second middle element. For example, if given linked list is 1->2->3->4->5->6 then the output should be 4.
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.
Method 2:
Traverse linked list using two pointers. Move one pointer by one and the other pointers by two. When the fast pointer reaches the end slow pointer will reach the middle of the linked list.
Below image shows how printMiddle function works in the code :
Python3
# Python3 program to find middle of # the linked list# Node class class Node: # Function to initialise the # node object def __init__(self, data): # Assign data self.data = data # Initialize next as null self.next = None # Linked List class contains a # Node object class LinkedList: # Function to initialize head 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 # Print the linked list def printList(self): node = self.head while node: print(str(node.data) + "->", end = "") node = node.next print("NULL") # Function that returns middle. def printMiddle(self): # Initialize two pointers, one will go # one step a time (slow), another two # at a time (fast) slow = self.head fast = self.head # Iterate till fast's next is null (fast # reaches end) while fast and fast.next: slow = slow.next fast = fast.next.next # Return the slow's data, which would be # the middle element. print("The middle element is ", slow.data)# Driver codeif __name__=='__main__': # Start with the empty list llist = LinkedList() for i in range(5, 0, -1): llist.push(i) llist.printList() llist.printMiddle()# This code is contributed by Kumar Shivam (kshivi99) |
Output:
5->NULL The middle element is [5] 4->5->NULL The middle element is [5] 3->4->5->NULL The middle element is [4] 2->3->4->5->NULL The middle element is [4] 1->2->3->4->5->NULL The middle element is [3]
Time Complexity: O(n) where n is the number of nodes in the given linked list.
Auxiliary Space: O(1), no extra space is required, so it is a constant.
Method 3:
Initialize mid element as head and initialize a counter as 0. Traverse the list from head, while traversing increment the counter and change mid to mid->next whenever the counter is odd. So the mid will move only half of the total length of the list.
Thanks to Narendra Kangralkar for suggesting this method.
Python3
# Python program to implement# the above approach# Node class class Node: # Function to initialise the # node object def __init__(self, data): # Assign data self.data = data # Initialize next as null self.next = None # Linked List class contains a # Node object class LinkedList: # Function to initialize head 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 # Print the linked list def printList(self): node = self.head while node: print(str(node.data) + "->", end = "") node = node.next print("NULL") # Function to get the middle of # the linked list def printMiddle(self): count = 0 mid = self.head heads = self.head while(heads != None): # Update mid, when 'count' # is odd number if count & 1: mid = mid.next count += 1 heads = heads.next # If empty list is provided if mid != None: print("The middle element is ", mid.data) # Driver codeif __name__=='__main__': # Start with the empty list llist = LinkedList() for i in range(5, 0, -1): llist.push(i) llist.printList() llist.printMiddle() # This code is contributed by Manisha_Ediga |
Output:
5->NULL The middle element is [5] 4->5->NULL The middle element is [5] 3->4->5->NULL The middle element is [4] 2->3->4->5->NULL The middle element is [4] 1->2->3->4->5->NULL The middle element is [3]
Time Complexity: O(n) where n is the number of nodes in the given linked list.
Auxiliary Space: O(1), no extra space is required, so it is a constant.
Please refer complete article on Find the middle of a given linked list for more details!
