How to Sort a LinkedList in Java?

Namachila 2 days ago

0 0 9 minutes read

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 Sort
 
public 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 Sort
 
public 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 List
 
public 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 Sort
 
public 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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