# Javascript Program For Sorting Linked List Which Is Already Sorted On Absolute Values

Given a linked list that is sorted based on absolute values. Sort the list based on actual values.**Examples:**

Input:1 -> -10Output:-10 -> 1Input:1 -> -2 -> -3 -> 4 -> -5Output:-5 -> -3 -> -2 -> 1 -> 4Input:-5 -> -10Output:-10 -> -5Input:5 -> 10Output:5 -> 10

Source : Amazon Interview

A simple solution is to traverse the linked list from beginning to end. For every visited node, check if it is out of order. If it is, remove it from its current position and insert it at the correct position. This is the implementation of insertion sort for linked list and the time complexity of this solution is O(n*n).

A better solution is to sort the linked list using merge sort. Time complexity of this solution is O(n Log n).

An efficient solution can work in O(n) time. An important observation is, all negative elements are present in reverse order. So we traverse the list, whenever we find an element that is out of order, we move it to the front of the linked list.

Below is the implementation of the above idea.

ever we find an element that is out of order, we move it to the front of the linked list.

Below is the implementation of the above idea.

## Javascript

`<script>` `// Javascript program to sort a linked list,` `// already sorted by absolute values` `// head of list` `var` `head;` `// Linked list Node` `class Node` `{` ` ` `constructor(d)` ` ` `{` ` ` `this` `.data = d;` ` ` `this` `.next = ` `null` `;` ` ` `}` `}` `// To sort a linked list by actual values.` `// The list is assumed to be sorted by` `// absolute values.` `function` `sortedList(head)` `{` ` ` `// Initialize previous and current` ` ` `// nodes` ` ` `var` `prev = head;` ` ` `var` `curr = head.next;` ` ` `// Traverse list` ` ` `while` `(curr != ` `null` `)` ` ` `{` ` ` `// If curr is smaller than prev, then` ` ` `// it must be moved to head` ` ` `if` `(curr.data < prev.data)` ` ` `{` ` ` `// Detach curr from linked list` ` ` `prev.next = curr.next;` ` ` `// Move current node to beginning` ` ` `curr.next = head;` ` ` `head = curr;` ` ` `// Update current` ` ` `curr = prev;` ` ` `}` ` ` `// Nothing to do if current element` ` ` `// is at right place` ` ` `else` ` ` `prev = curr;` ` ` `// Move current` ` ` `curr = curr.next;` ` ` `}` ` ` `return` `head;` `}` `/* Inserts a new Node at front of` ` ` `the list. */` `function` `push(new_data)` `{` ` ` `/* 1 & 2: Allocate the Node &` ` ` `Put in the data */` ` ` `var` `new_node = ` `new` `Node(new_data);` ` ` `// 3. Make next of new Node as head` ` ` `new_node.next = head;` ` ` `// 4. Move the head to point to` ` ` `// new Node` ` ` `head = new_node;` `}` `// Function to print linked list` `function` `printList(head)` `{` ` ` `var` `temp = head;` ` ` `while` `(temp != ` `null` `)` ` ` `{` ` ` `document.write(temp.data + ` `" "` `);` ` ` `temp = temp.next;` ` ` `}` ` ` `document.write(` `"<br/>"` `);` `}` `// Driver code ` `/* Constructed Linked List is` ` ` `1->2->3->4->5->6-> 7->8->8->` ` ` `9->null */` `push(-5);` `push(5);` `push(4);` `push(3);` `push(-2);` `push(1);` `push(0);` `document.write(` ` ` `"Original List :<br/>"` `);` `printList(head);` `head = sortedList(head);` `document.write(` ` ` `"Sorted list :<br/>"` `);` `printList(head);` `// This code is contributed by aashish1995` `</script>` |

**Output:**

Original list : 0 -> 1 -> -2 -> 3 -> 4 -> 5 -> -5 Sorted list : -5 -> -2 -> 0 -> 1 -> 3 -> 4 -> 5

**Time Complexity: **O(N)

**Auxiliary Space: **O(1)

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