Given a singly linked list and a position, delete a linked list node at the given position.
Example:
Input: position = 1, Linked List = 8->2->3->1->7 Output: Linked List = 8->3->1->7 Input: position = 0, Linked List = 8->2->3->1->7 Output: Linked List = 2->3->1->7
If the node to be deleted is the root, simply delete it. To delete a middle node, we must have a pointer to the node previous to the node to be deleted. So if positions are not zero, we run a loop position-1 times and get a pointer to the previous node.
Below is the implementation of the above idea.
Python
# Python program to delete a node in a linked list# at a given position# Node class class Node: # Constructor to initialize the node object def __init__(self, data): self.data = data self.next = Noneclass LinkedList: # Constructor 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 # Given a reference to the head of a list # and a position, delete the node at a given position def deleteNode(self, position): # If linked list is empty if self.head == None: return # Store head node temp = self.head # If head needs to be removed if position == 0: self.head = temp.next temp = None return # Find previous node of the node to be deleted for i in range(position -1 ): temp = temp.next if temp is None: break # If position is more than number of nodes if temp is None: return if temp.next is None: return # Node temp.next is the node to be deleted # store pointer to the next of node to be deleted next = temp.next.next # Unlink the node from linked list temp.next = None temp.next = next # Utility function to print the linked LinkedList def printList(self): temp = self.head while(temp): print " %d " %(temp.data), temp = temp.next# Driver program to test above functionllist = LinkedList()llist.push(7)llist.push(1)llist.push(3)llist.push(2)llist.push(8)print "Created Linked List: "llist.printList()llist.deleteNode(4)print "Linked List after Deletion at position 4: "llist.printList()# This code is contributed by Nikhil Kumar Singh(nickzuck_007) |
Output:
Created Linked List: 8 2 3 1 7 Linked List after Deletion at position 4: 8 2 3 1
Time Complexity: O(n), where n represents the length of the given linked list.
Auxiliary Space: O(1), no extra space is required, so it is a constant.
Please refer complete article on Delete a Linked List node at a given position for more details!
