Prerequisite: Merge Sort for Linked Lists Merge sort is often preferred for sorting a linked list. The slow random-access performance of a linked list makes some other algorithms (such as quicksort) perform poorly, and others (such as heapsort) completely impossible.
In this post, Merge sort for linked list is implemented using JavaScript.
Examples:
Input : 5 -> 4 -> 3 -> 2 -> 1
Output :1 -> 2 -> 3 -> 4 -> 5Input : 10 -> 20 -> 3 -> 2 -> 1
Output : 1 -> 2 -> 3 -> 10 -> 20
Implementation:
javascript
<script> // Create Node of LinkedList function Node(data) { this.node = data; this.next = null; } // To initialize a linkedlist function LinkedList(list) { this.head = list || null } // Function to insert The new Node into the linkedList LinkedList.prototype.insert = function(data) { // Check if the linked list is empty // so insert first node and lead head // points to generic node if (this.head === null) this.head = new Node(data); else { // If linked list is not empty, insert the node // at the end of the linked list let list = this.head; while (list.next) { list = list.next; } // Now here list pointer points to last // node let’s insert out new node in it list.next = new Node(data) } } // Function to print linkedList LinkedList.prototype.iterate = function() { // First we will check whether out // linked list is empty or node if (this.head === null) return null; // If linked list is not empty we will // iterate from each Node and prints // it’s value store in “data” property let list = this.head; // we will iterate until our list variable // contains the “Next” value of the last Node // i.e-> null while (list) { document.write(list.node) if (list.next) document.write(' -> ') list = list.next } } // Function to mergesort a linked list LinkedList.prototype.mergeSort = function(list) { if (list.next === null) return list; let count = 0; let countList = list let leftPart = list; let leftPointer = list; let rightPart = null; let rightPointer = null; // Counting the nodes in the received linkedlist while (countList.next !== null) { count++; countList = countList.next; } // counting the mid of the linked list let mid = Math.floor(count / 2) let count2 = 0; // separating the left and right part with // respect to mid node in the linked list while (count2 < mid) { count2++; leftPointer = leftPointer.next; } rightPart = new LinkedList(leftPointer.next); leftPointer.next = null; // Here are two linked list which // contains the left most nodes and right // most nodes of the mid node return this._mergeSort(this.mergeSort(leftPart), this.mergeSort(rightPart.head)) } // Merging both lists in a sorted manner LinkedList.prototype._mergeSort = function(left, right) { // Create a new empty linked list let result = new LinkedList() let resultPointer = result.head; let pointerLeft = left; let pointerRight = right; // If true then add left most node value in result, // increment left pointer else do the same in // right linked list. // This loop will be executed until pointer's of // a left node or right node reached null while (pointerLeft && pointerRight) { let tempNode = null; // Check if the right node's value is greater than // left node's value if (pointerLeft.node > pointerRight.node) { tempNode = pointerRight.node pointerRight = pointerRight.next; } else { tempNode = pointerLeft.node pointerLeft = pointerLeft.next; } if (result.head == null) { result.head = new Node(tempNode) resultPointer = result.head } else { resultPointer.next = new Node(tempNode) resultPointer = resultPointer.next } } // Add the remaining elements in the last of resultant // linked list resultPointer.next = pointerLeft; while (resultPointer.next) resultPointer = resultPointer.next resultPointer.next = pointerRight // Result is the new sorted linked list return result.head; } // Initialize the object let l = new LinkedList(); l.insert(10) l.insert(20) l.insert(3) l.insert(2) l.insert(1) // Print the linked list l.iterate() // Sort the linked list l.head = LinkedList.prototype.mergeSort(l.head) document.write('<br> After sorting : '); // Print the sorted linked list l.iterate() </script> |
Output
10 -> 20 -> 3 -> 2 -> 1 After sorting : 1 -> 2 -> 3 -> 10 -> 20
Complexity Analysis:
- Time Complexity: O(n*log(n))
- Auxiliary Space: O(n)
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