C++ Program To Delete Middle Of Linked List
Last Updated :
30 Dec, 2021
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.Â
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.Â
C++14
#include <bits/stdc++.h>
using namespace std;
struct Node
{
int data;
struct Node* next;
};
int countOfNodes( struct Node* head)
{
int count = 0;
while (head != NULL)
{
head = head->next;
count++;
}
return count;
}
struct Node* deleteMid( struct Node* head)
{
if (head == NULL)
return NULL;
if (head->next == NULL)
{
delete head;
return NULL;
}
struct Node* copyHead = head;
int count = countOfNodes(head);
int mid = count / 2;
while (mid-- > 1)
{
head = head->next;
}
head->next = head->next->next;
return copyHead;
}
void printList( struct Node* ptr)
{
while (ptr != NULL)
{
cout << ptr->data << "->" ;
ptr = ptr->next;
}
cout << "NULL" ;
}
Node* newNode( int data)
{
struct Node* temp = new Node;
temp->data = data;
temp->next = NULL;
return temp;
}
int main()
{
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:Â
- Time Complexity: O(n).Â
Two traversals of the linked list is needed
- Auxiliary Space: O(1).Â
No extra space is needed.
Efficient solution:Â
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. Â
C++
#include <bits/stdc++.h>
using namespace std;
struct Node
{
int data;
struct Node* next;
};
struct Node* deleteMid( struct Node* head)
{
if (head == NULL)
return NULL;
if (head->next == NULL)
{
delete head;
return NULL;
}
struct Node* slow_ptr = head;
struct Node* fast_ptr = head;
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;
}
prev->next = slow_ptr->next;
delete slow_ptr;
return head;
}
void printList( struct Node* ptr)
{
while (ptr != NULL)
{
cout << ptr->data << "->" ;
ptr = ptr->next;
}
cout << "NULL" ;
}
Node* newNode( int data)
{
struct Node* temp = new Node;
temp->data = data;
temp->next = NULL;
return temp;
}
int main()
{
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:Â
- Time Complexity: O(n).Â
Only one traversal of the linked list is needed
- Auxiliary Space: O(1).Â
As no extra space is needed.
Please refer complete article on Delete middle of linked list for more details!
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