A Linked List is a linear data structure, in which the elements are not stored at contiguous memory locations.
Sorting the nodes of a Singly Linked list in ascending order:
Original List
Sorted List
We can sort the LinkedList by many sorting techniques:
Method 1: Sort Linked List using Bubble Sort
- To accomplish this task, we maintain two pointers: current and index.
- Initially, current point to head node and index will point to node next to current.
- Traverse through the list till current points to null, by comparing current’s data with index’s data.
- And for each current’s value, index is the next to current node which traverse from current’s next node till null.
- And then the value of current node is compared with every value from its next node till last and if the value is smaller than the current value, then the values are swapped and in this way the least value comes as current index.
Java
// Java program to sort a Linked List using Bubble Sortpublic class SortList { // Represent a node of the singly linked list class Node { int data; Node next; public Node(int data) { this.data = data; this.next = null; } } // Represent the head and tail of the singly linked list public Node head = null; public Node tail = null; // addNode() will add a new node to the list public void addNode(int data) { // Create a new node Node newNode = new Node(data); // Checks if the list is empty if (head == null) { // If list is empty, both head and tail will // point to new node head = newNode; tail = newNode; } else { // newNode will be added after tail such that // tail's next will point to newNode tail.next = newNode; // newNode will become new tail of the list tail = newNode; } } // sortList() will sort nodes of the list in ascending // order public void sortList() { // Node current will point to head Node current = head, index = null; int temp; if (head == null) { return; } else { while (current != null) { // Node index will point to node next to // current index = current.next; while (index != null) { // If current node's data is greater // than index's node data, swap the data // between them if (current.data > index.data) { temp = current.data; current.data = index.data; index.data = temp; } index = index.next; } current = current.next; } } } // display() will display all the nodes present in the // list public void display() { // Node current will point to head Node current = head; if (head == null) { System.out.println("List is empty"); return; } while (current != null) { // Prints each node by incrementing pointer System.out.print(current.data + " "); current = current.next; } System.out.println(); } public static void main(String[] args) { SortList sList = new SortList(); // Adds data to the list sList.addNode(8); sList.addNode(3); sList.addNode(7); sList.addNode(4); // Displaying original list System.out.println("Original list: "); sList.display(); // Sorting list sList.sortList(); // Displaying sorted list System.out.println("Sorted list: "); sList.display(); }} |
Output
Original list: 8 3 7 4 Sorted list: 3 4 7 8
Time complexity: O(n ^ 2)
Auxiliary Space: O(1)
Method 2: Sort Linked List using Insertion Sort
- In the Insertion sort technique, we assume that all the elements before the current element in the list is already sorted, and we begin with the current element.
- The current element is compared with all the elements before it and swapped if not in order. This process is repeated for all the subsequent elements.
- In general, the Insertion sort technique compares each element with all of its previous elements and sorts the element to place it in its proper position.
As already mentioned, the Insertion sort technique is more feasible for a smaller set of data, and thus arrays with a few elements can be sorted using efficiently Insertion sort.
Insertion sort is especially useful in sorting linked list data structures. As you know, Linked lists have pointers pointing to its next element (singly linked list) and previous element (double linked list). This makes it easier to keep track of the previous and next elements.
Java
// Java program to sort Linked List using Insertion Sortpublic class LinkedlistIS { node head; node sorted; class node { int val; node next; public node(int val) { this.val = val; } } void push(int val) { // allocate node node newnode = new node(val); // link the old list of the new node newnode.next = head; // move the head to point to the new node head = newnode; } // function to sort a singly linked list using insertion // sort void insertionSort(node headref) { // Initialize sorted linked list sorted = null; node current = headref; // Traverse the given linked list and insert every // node to sorted while (current != null) { // Store next for next iteration node next = current.next; // insert current in sorted linked list sortedInsert(current); // Update current current = next; } // Update head_ref to point to sorted linked list head = sorted; } // function to insert a new_node in a list. Note that // this function expects a pointer to head_ref as this // can modify the head of the input linked list // (similar to push()) void sortedInsert(node newnode) { // Special case for the head end if (sorted == null || sorted.val >= newnode.val) { newnode.next = sorted; sorted = newnode; } else { node current = sorted; // Locate the node before the point of insertion while (current.next != null && current.next.val < newnode.val) { current = current.next; } newnode.next = current.next; current.next = newnode; } } // Function to print linked list void printlist(node head) { while (head != null) { System.out.print(head.val + " "); head = head.next; } } // Driver program to test above functions public static void main(String[] args) { LinkedlistIS list = new LinkedlistIS(); list.push(4); list.push(7); list.push(3); list.push(8); System.out.println("Linked List before Sorting.."); list.printlist(list.head); list.insertionSort(list.head); System.out.println("\nLinkedList After sorting"); list.printlist(list.head); }} |
Output
Linked List before Sorting.. 8 3 7 4 LinkedList After sorting 3 4 7 8
Time complexity: O(n ^ 2)
Auxiliary Space: O(1)
Method 3: Sort Linked List using Quick Sort
Quick sort follows divide and conquer approach. It picks an element as pivot and partitions the given array around the picked pivot.
The key process in quickSort is partition(). Target of partitions is, given an array and an element x of array as pivot, put x at its correct position in sorted array and put all smaller elements (smaller than x) before x, and put all greater elements (greater than x) after x. All this should be done in linear time.
Quick sort is preferred over merge sort as Quick sort is an in-place algorithm (meaning, no additional memory space required).
Java
// Java program for Quick Sort on Singly Linked Listpublic class QuickSortLinkedList { static class Node { int data; Node next; Node(int d) { this.data = d; this.next = null; } } Node head; void addNode(int data) { if (head == null) { head = new Node(data); return; } Node curr = head; while (curr.next != null) curr = curr.next; Node newNode = new Node(data); curr.next = newNode; } void printList(Node n) { while (n != null) { System.out.print(n.data); System.out.print(" "); n = n.next; } } // takes first and last node, // but do not break any links in // the whole linked list Node paritionLast(Node start, Node end) { if (start == end || start == null || end == null) return start; Node pivot_prev = start; Node curr = start; int pivot = end.data; // iterate till one before the end, // no need to iterate till the end // because end is pivot while (start != end) { if (start.data < pivot) { // keep tracks of last modified item pivot_prev = curr; int temp = curr.data; curr.data = start.data; start.data = temp; curr = curr.next; } start = start.next; } // swap the position of curr i.e. // next suitable index and pivot int temp = curr.data; curr.data = pivot; end.data = temp; // return one previous to current // because current is now pointing to pivot return pivot_prev; } void sort(Node start, Node end) { if (start == end) return; // split list and partition recurse Node pivot_prev = paritionLast(start, end); sort(start, pivot_prev); // if pivot is picked and moved to the start, // that means start and pivot is same // so pick from next of pivot if (pivot_prev != null && pivot_prev == start) sort(pivot_prev.next, end); // if pivot is in between of the list, // start from next of pivot, // since we have pivot_prev, so we move two nodes else if (pivot_prev != null && pivot_prev.next != null) sort(pivot_prev.next.next, end); } // Driver Code public static void main(String[] args) { QuickSortLinkedList list = new QuickSortLinkedList(); list.addNode(8); list.addNode(3); list.addNode(7); list.addNode(4); Node n = list.head; while (n.next != null) n = n.next; System.out.println("Original List: "); list.printList(list.head); list.sort(list.head, n); System.out.println("\nSorted List: "); list.printList(list.head); }} |
Output
Original List: 8 3 7 4 Sorted List: 3 4 7 8
Time complexity: O(n ^ 2)
Auxiliary Space: O(1)
Method 3: Sort Linked List using Merge Sort
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.
Merge Sort is a Divide and Conquer algorithm. It divides the input array into two halves, calls itself for the two halves, and then merges the two sorted halves. The merge() function is used for merging two halves. The merge(arr, l, m, r) is a key process that assumes that arr[l..m] and arr[m+1..r] are sorted and merges the two sorted sub-arrays into one.
- Let head be the first node of the linked list to be sorted and headRef be the pointer to head.
- Note that we need a reference to head in MergeSort() as the below implementation changes next links to sort the linked lists (not data at the nodes), so head node has to be changed if the data at original head is not the smallest value in linked list.
Java
// Java program to sort linkedList using Merge Sortpublic class linkedList { node head = null; // node a, b; static class node { int val; node next; public node(int val) { this.val = val; } } node sortedMerge(node a, node b) { node result = null; // Base cases if (a == null) return b; if (b == null) return a; // Pick either a or b, and recur if (a.val < b.val) { result = a; result.next = sortedMerge(a.next, b); } else { result = b; result.next = sortedMerge(a, b.next); } return result; } node mergeSort(node h) { // Base case : if head is null if (h == null || h.next == null) { return h; } // get the middle of the list node middle = getMiddle(h); node nextofmiddle = middle.next; // set the next of middle node to null middle.next = null; // Apply mergeSort on left list node left = mergeSort(h); // Apply mergeSort on right list node right = mergeSort(nextofmiddle); // Merge the left and right lists node sortedlist = sortedMerge(left, right); return sortedlist; } // Utility function to get the middle of the linked list public static node getMiddle(node head) { if (head == null) return head; node slow = head, fast = head; while (fast.next != null && fast.next.next != null) { slow = slow.next; fast = fast.next.next; } return slow; } void push(int new_data) { // allocate node node new_node = new node(new_data); // link the old list of the new node new_node.next = head; // move the head to point to the new node head = new_node; } // Utility function to print the linked list void printList(node headref) { while (headref != null) { System.out.print(headref.val + " "); headref = headref.next; } } public static void main(String[] args) { linkedList li = new linkedList(); li.push(4); li.push(7); li.push(3); li.push(8); System.out.print("\nOriginal List: \n"); li.printList(li.head); // Apply merge Sort li.head = li.mergeSort(li.head); System.out.print("\nSorted List: \n"); li.printList(li.head); }} |
Output
Original List: 8 3 7 4 Sorted List: 3 4 7 8
Time complexity: O(n log n)
Auxiliary Space: O(1)
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