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.
// C++ program to delete middle // of a linked list #include <bits/stdc++.h> using namespace std;
// Link list Node struct Node
{ int data;
struct Node* next;
}; // Count of nodes int countOfNodes( struct Node* head)
{ int count = 0;
while (head != NULL)
{
head = head->next;
count++;
}
return count;
} // Deletes middle node and returns // head of the modified list struct Node* deleteMid( struct Node* head)
{ // Base cases
if (head == NULL)
return NULL;
if (head->next == NULL)
{
delete head;
return NULL;
}
struct Node* copyHead = head;
// Find the count of nodes
int count = countOfNodes(head);
// Find the middle node
int mid = count / 2;
// Delete the middle node
while (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 void printList( struct Node* ptr)
{ while (ptr != NULL)
{
cout << ptr->data << "->" ;
ptr = ptr->next;
}
cout << "NULL" ;
} // Utility function to create // a new node. Node* newNode( int data)
{ struct Node* temp = new Node;
temp->data = data;
temp->next = NULL;
return temp;
} // Driver code int main()
{ // Start with the empty list
struct Node* head = newNode(1);
head->next = newNode(2);
head->next->next = newNode(3);
head->next->next->next = newNode(4);
cout << "Given Linked List" ;
printList(head);
head = deleteMid(head);
cout << "Linked List after deletion of middle" ;
printList(head);
return 0;
} |
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.
// C++ program to delete middle // of a linked list #include <bits/stdc++.h> using namespace std;
// Link list Node struct Node
{ int data;
struct Node* next;
}; // Deletes middle node and returns // head of the modified list struct Node* deleteMid( struct Node* head)
{ // Base cases
if (head == NULL)
return NULL;
if (head->next == NULL)
{
delete head;
return NULL;
}
// Initialize slow and fast pointers
// to reach middle of linked list
struct Node* slow_ptr = head;
struct Node* fast_ptr = head;
// Find the middle and previous
// of middle.
// To store previous of slow_ptr
struct Node* prev;
while (fast_ptr != NULL &&
fast_ptr->next != NULL)
{
fast_ptr = fast_ptr->next->next;
prev = slow_ptr;
slow_ptr = slow_ptr->next;
}
// Delete the middle node
prev->next = slow_ptr->next;
delete slow_ptr;
return head;
} // A utility function to print // a given linked list void printList( struct Node* ptr)
{ while (ptr != NULL)
{
cout << ptr->data << "->" ;
ptr = ptr->next;
}
cout << "NULL" ;
} // Utility function to create // a new node. Node* newNode( int data)
{ struct Node* temp = new Node;
temp->data = data;
temp->next = NULL;
return temp;
} // Driver code int main()
{ // Start with the empty list
struct Node* head = newNode(1);
head->next = newNode(2);
head->next->next = newNode(3);
head->next->next->next = newNode(4);
cout << "Given Linked List" ;
printList(head);
head = deleteMid(head);
cout << "Linked List after deletion of middle" ;
printList(head);
return 0;
} |
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!