Given a linked list containing n nodes. The problem is to insert a new node with data x at the middle of the list. If n is even, then insert the new node after the (n/2)th node, else insert the new node after the (n+1)/2th node.
Examples:
Input : list: 1->2->4->5 x = 3 Output : 1->2->3->4->5 Input : list: 5->10->4->32->16 x = 41 Output : 5->10->4->41->32->16
Method 1(Using length of the linked list):
Find the number of nodes or length of the linked using one traversal. Let it be len. Calculate c = (len/2), if len is even, else c = (len+1)/2, if len is odd. Traverse again the first c nodes and insert the new node after the cth node.
Javascript
<script> // Javascript implementation to insert node // at the middle of the linked list var head; // head of list /* Node Class */ class Node { // Constructor to create a new node constructor(d) { this .data = d; this .next = null ; } } // function to insert node at the // middle of the linked list function insertAtMid(x) { // if list is empty if (head == null ) head = new Node(x); else { // get a new node var newNode = new Node(x); var ptr = head; var len = 0; // calculate length of the linked list // , i.e, the number of nodes while (ptr != null ) { len++; ptr = ptr.next; } // 'count' the number of nodes after which // the new node is to be inserted var count = ((len % 2) == 0) ? (len / 2) : (len + 1) / 2; ptr = head; // 'ptr' points to the node after which // the new node is to be inserted while (count-- > 1) ptr = ptr.next; // insert the 'newNode' and adjust // the required links newNode.next = ptr.next; ptr.next = newNode; } } // function to display the linked list function display() { var temp = head; while (temp != null ) { document.write(temp.data + " " ); temp = temp.next; } } // Driver program to test above // Creating the list 1.2.4.5 head = new Node(1); head.next = new Node(2); head.next.next = new Node(4); head.next.next.next = new Node(5); document.write( "Linked list before " + "insertion: " ); display(); var x = 3; insertAtMid(x); document.write( "<br/>Linked list after" + " insertion: " ); display(); // This code contributed by Rajput-Ji </script> |
Output:
Linked list before insertion: 1 2 4 5 Linked list after insertion: 1 2 3 4 5
Time Complexity : O(n)
Auxiliary Space : O(1)
Method 2(Using two pointers):
Based on the tortoise and hare algorithm which uses two pointers, one known as slow and the other known as fast. This algorithm helps in finding the middle node of the linked list. It is explained in the front and black split procedure of this post. Now, you can insert the new node after the middle node obtained from the above process. This approach requires only a single traversal of the list.
Javascript
<script> // Javascript implementation to insert node // at the middle of the linked list var head; // head of list /* Node Class */ class Node { // Constructor to create a new node constructor(val) { this .data = val; this .next = null ; } } // function to insert node at the // middle of the linked list function insertAtMid(x) { // if list is empty if (head == null ) head = new Node(x); else { // get a new node var newNode = new Node(x); // assign values to the slow // and fast pointers var slow = head; var fast = head.next; while (fast != null && fast.next != null ) { // move slow pointer to next node slow = slow.next; // move fast pointer two nodes // at a time fast = fast.next.next; } // insert the 'newNode' and adjust // the required links newNode.next = slow.next; slow.next = newNode; } } // function to display the linked list function display() { var temp = head; while (temp != null ) { document.write(temp.data + " " ); temp = temp.next; } } // Driver program to test above // Creating the list 1.2.4.5 head = null ; head = new Node(1); head.next = new Node(2); head.next.next = new Node(4); head.next.next.next = new Node(5); document.write( "Linked list before" + " insertion: " ); display(); var x = 3; insertAtMid(x); document.write( "<br/>Linked list after" + " insertion: " ); display(); // This code is contributed by todaysgaurav </script> |
Output:
Linked list before insertion: 1 2 4 5 Linked list after insertion: 1 2 3 4 5
Time Complexity: O(n)
Space complexity: O(n) where n is size of linked list
Please refer complete article on Insert node into the middle of the linked list for more details!
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