Javascript Program For Rearranging A Given Linked List In-Place
Last Updated : 05 May, 2025
Given a singly linked list L0 -> L1 -> … -> Ln-1 -> Ln. Rearrange the nodes in the list so that the new formed list is : L0 -> Ln -> L1 -> Ln-1 -> L2 -> Ln-2 ...
You are required to do this in place without altering the nodes' values.
Examples:
Input: 1 -> 2 -> 3 -> 4
Output: 1 -> 4 -> 2 -> 3
Input: 1 -> 2 -> 3 -> 4 -> 5
Output: 1 -> 5 -> 2 -> 4 -> 3
Simple Solution:
1) Initialize current node as head.
2) While next of current node is not null, do following
a) Find the last node, remove it from the end and insert it as next
of the current node.
b) Move current to next to next of current
The time complexity of the above simple solution is O(n2) where n is the number of nodes in the linked list.
Better Solution:
- Copy contents of the given linked list to a vector.
- Rearrange the given vector by swapping nodes from both ends.
- Copy the modified vector back to the linked list.
Efficient Solution:
1) Find the middle point using tortoise and hare method.
2) Split the linked list into two halves using found middle point in step 1.
3) Reverse the second half.
4) Do alternate merge of first and second halves.
The Time Complexity of this solution is O(n).
Below is the implementation of this method.
JavaScript // Javascript program to rearrange // linked list in place // Linked List Class // Head of the list let head; // Node Class class Node { // Constructor to create // a new node constructor(d) { this.data = d; this.next = null; } } function printlist(node) { if (node == null) { return; } while (node != null) { console.log(node.data); node = node.next; } } function reverselist(node) { let prev = null, curr = node, next; while (curr != null) { next = curr.next; curr.next = prev; prev = curr; curr = next; } node = prev; return node; } function rearrange(node) { // 1) Find the middle point using // tortoise and hare method let slow = node, fast = slow.next; while (fast != null && fast.next != null) { slow = slow.next; fast = fast.next.next; } // 2) Split the linked list in // two halves // node1, head of first half // 1 -> 2 -> 3 // node2, head of second half // 4 -> 5 let node1 = node; let node2 = slow.next; slow.next = null; // 3) Reverse the second half, // i.e., 5 -> 4 node2 = reverselist(node2); // 4) Merge alternate nodes // Assign dummy Node node = new Node(0); // curr is the pointer to this // dummy Node, which will be // used to form the new list let curr = node; while (node1 != null || node2 != null) { // First add the element from // first list if (node1 != null) { curr.next = node1; curr = curr.next; node1 = node1.next; } // Then add the element from // second list if (node2 != null) { curr.next = node2; curr = curr.next; node2 = node2.next; } } // Assign the head of the new // list to head pointer node = node.next; } head = new Node(1); head.next = new Node(2); head.next.next = new Node(3); head.next.next.next = new Node(4); head.next.next.next.next = new Node(5); // Print original list console.log("Original Linked List"); printlist(head); // Rearrange list as per ques console.log("Rearrange Linked List"); rearrange(head); // Print modified list printlist(head); // This code is contributed by gauravrajput1 OutputOriginal Linked List 1 2 3 4 5 Rearrange Linked List 1 5 2 4 3
Complexity Analysis:
- Time Complexity: O(n)
- Auxiliary Space: O(1)
Thanks to Gaurav Ahirwar for suggesting the above approach.
Another approach:
1. Take two pointers prev and curr, which hold the addresses of head and head-> next.
2. Compare their data and swap.
After that, a new linked list is formed.
Below is the implementation:
JavaScript // Javascript code to rearrange linked list // in place class Node { constructor() { this.data; this.next = null; } } // Function for rearranging a linked list // with high and low value. function rearrange(head) { // Base case if (head == null) return null; // Two pointer variable. let prev = head, curr = head.next; while (curr != null) { // Swap function for swapping data. if (prev.data > curr.data) { let t = prev.data; prev.data = curr.data; curr.data = t; } // Swap function for swapping data if (curr.next != null && curr.next.data > curr.data) { let t = curr.next.data; curr.next.data = curr.data; curr.data = t; } prev = curr.next; if (curr.next == null) break; curr = curr.next.next; } return head; } // Function to display Node // of linked list. function display(head) { let curr = head; while (curr != null) { console.log(curr.data); curr = curr.next; } } // Function to insert a Node in // the linked list at the beginning. function push(head, k) { let tem = new Node(); tem.data = k; tem.next = head; head = tem; return head; } // Driver code let head = null; head = push(head, 7); head = push(head, 3); head = push(head, 8); head = push(head, 6); head = push(head, 9); head = rearrange(head); display(head); // This code is contributed by unknown2108 Complexity Analysis:
- Time Complexity : O(n)
- Auxiliary Space : O(1)
Another Approach: (Using recursion)
- Hold a pointer to the head node and go till the last node using recursion
- Once the last node is reached, start swapping the last node to the next of head node
- Move the head pointer to the next node
- Repeat this until the head and the last node meet or come adjacent to each other
- Once the Stop condition met, we need to discard the left nodes to fix the loop created in the list while swapping nodes.
JavaScript // Javascript program to implement // the above approach // Creating the structure for node class Node { // Function to create newNode // in a linkedlist constructor(val) { this.data = val; this.next = null; } } letleft = null; // Function to print the list function printlist(head) { while (head != null) { console.log(head.data); if (head.next != null) { } head = head.next; } } // Function to rearrange function rearrange(head) { if (head != null) { left = head; reorderListUtil(left); } } function reorderListUtil(right) { if (right == null) { return; } reorderListUtil(right.next); // We set left = null, when we reach // stop condition, so no processing // required after that if (left == null) { return; } // Stop condition: odd case : // left = right, even // case : left.next = right if (left != right && left.next != right) { let temp = left.next; left.next = right; right.next = temp; left = temp; } else { // Stop condition , set null // to left nodes if (left.next == right) { // even case left.next.next = null; left = null; } else { // odd case left.next = null; left = null; } } } // Drivers Code let head = new Node(1); head.next = new Node(2); head.next.next = new Node(3); head.next.next.next = new Node(4); head.next.next.next.next = new Node(5); // Print original list console.log("Original Linked List: "); printlist(head); // Modify the list console.log("Rearranged Linked List: "); rearrange(head); // Print modified list printlist(head); // This code is contributed by aashish1995 OutputOriginal Linked List: 1 2 3 4 5 Rearranged Linked List: 1 5 2 4 3
Complexity Analysis:
- Time Complexity: O(n), where n represents the length of the given linked list.
- Auxiliary Space: O(n), due to recursive call stack where n represents the length of the given linked list.
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