How to Sort a LinkedList in Java?

03 Jul, 2022
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:
Bubble sortInsertion sortQuick sortMerge sort
Bubble sort
Insertion sort
Quick sort
Merge sort
Method 1: Sort 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();    }}
Original list: 
8 3 7 4 
Sorted list: 
3 4 7 8
Time complexity: O(n ^ 2)
Space complexity: O(1)
Method 2: Sort 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 off 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);     } } 
Linked List before Sorting..
8 3 7 4 
LinkedList After sorting
3 4 7 8
Time complexity: O(n ^ 2)
Space complexity: O(1)
Method 3: Sort 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); } }
Original List: 
8 3 7 4 
Sorted List: 
3 4 7 8
Time complexity: O(n ^ 2)
Space complexity: O(1)
Method 3: Sort 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 off 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);     } } 
Original List: 
8 3 7 4 
Sorted List: 
3 4 7 8
Time complexity: O(n log n)
Space complexity: O(1)
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