Given a singly linked list, delete the middle of the linked list. For example, if the given linked list is 1->2->3->4->5 then the linked list should be modified to 1->2->4->5
If there are even nodes, then there would be two middle nodes, we need to delete the second middle element. For example, if given linked list is 1->2->3->4->5->6 then it should be modified to 1->2->3->5->6.
If the input linked list is NULL, then it should remain NULL.
If the input linked list has 1 node, then this node should be deleted and a new head should be returned.
Simple solution: The idea is to first count the number of nodes in a linked list, then delete n/2’th node using the simple deletion process.
Python3
# Python3 program to delete middle # of a linked list # Link list Node class Node: def __init__(self): self.data = 0 self.next = None # Count of nodes def countOfNodes(head): count = 0 while (head != None): head = head.next count += 1 return count # Deletes middle node and returns # head of the modified list def deleteMid(head): # Base cases if (head == None): return None if (head.next == None): del head return None copyHead = head # Find the count of nodes count = countOfNodes(head) # Find the middle node mid = count // 2 # Delete the middle node while (mid > 1): mid -= 1 head = head.next # Delete the middle node head.next = head.next.next return copyHead # A utility function to print # a given linked list def printList(ptr): while (ptr != None): print(ptr.data, end = '->') ptr = ptr.next print('NULL') # Utility function to create # a new node. def newNode(data): temp = Node() temp.data = data temp.next = None return temp # Driver Code if __name__=='__main__': # Start with the empty list head = newNode(1) head.next = newNode(2) head.next.next = newNode(3) head.next.next.next = newNode(4) print("Given Linked List") printList(head) head = deleteMid(head) print("Linked List after deletion of middle") printList(head) # This code is contributed by rutvik_56 |
Output:
Given Linked List 1->2->3->4->NULL Linked List after deletion of middle 1->2->4->NULL
Complexity Analysis:
- Time Complexity: O(n).
Two traversals of the linked list is needed - Auxiliary Space: O(1).
No extra space is needed.
Efficient solution:
Approach: The above solution requires two traversals of the linked list. The middle node can be deleted using one traversal. The idea is to use two pointers, slow_ptr, and fast_ptr. Both pointers start from the head of list. When fast_ptr reaches the end, slow_ptr reaches middle. This idea is same as the one used in method 2 of this post. The additional thing in this post is to keep track of the previous middle so the middle node can be deleted.
Below is the implementation.
Python3
# Python3 program to delete the # middle of a linked list # Linked List Node class Node: def __init__(self, data): self.data = data self.next = None # Create and handle list # operations class LinkedList: def __init__(self): # Head of the list self.head = None # Add new node to the list end def addToList(self, data): newNode = Node(data) if self.head is None: self.head = newNode return last = self.head while last.next: last = last.next last.next = newNode # Returns the list in string # format def __str__(self): linkedListStr = "" temp = self.head while temp: linkedListStr += str(temp.data) + "->" temp = temp.next return linkedListStr + "NULL" # Method deletes middle node def deleteMid(self): # Base cases if (self.head is None or self.head.next is None): return # Initialize slow and fast pointers # to reach middle of linked list slow_Ptr = self.head fast_Ptr = self.head # Find the middle and previous of # middle prev = None # To store previous of slow pointer while (fast_Ptr is not None and fast_Ptr.next is not None): fast_Ptr = fast_Ptr.next.next prev = slow_Ptr slow_Ptr = slow_Ptr.next # Delete the middle node prev.next = slow_Ptr.next # Driver code linkedList = LinkedList() linkedList.addToList(1) linkedList.addToList(2) linkedList.addToList(3) linkedList.addToList(4) print("Given Linked List") print(linkedList) linkedList.deleteMid() print("Linked List after deletion of middle") print(linkedList) # This code is contributed by Debidutta Rath |
Output:
Given Linked List 1->2->3->4->NULL Linked List after deletion of middle 1->2->4->NULL
Complexity Analysis:
- Time Complexity: O(n).
Only one traversal of the linked list is needed - Auxiliary Space: O(1).
As no extra space is needed.
Please refer complete article on Delete middle of linked list for more details!
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