Related questions: Sort Linked List using Merge Sort, Given a Linked list, Sort it using Merge Sort Algorithm.

Merge sort is preferred algorithm for sorting a linked list, lets see why,
The way linked list is structured which doesn't allow random access makes some other algorithms like Quicksort perform poorly, So Merge sort is often preferred algorithm for sorting linked list.

Lets understand what is the input and the expected output.

Algorithm (We will use Merge Sort for sorting Linked list.)
For sorting Linked list, we can use any algorithm but the most suitable algorithm is Merge Sort.

Why is merge sort preferred over quick sort for sorting linked lists?

As Singly Linked list can be accessed in only one direction and cannot be accessed randomly, Quick sort will not be well suitable here.

Quick sort requires access to elements in both direction for swapping and doing such operation in Linked list is not as easy as working with Arrays.
Starting from the end and moving backward is usually expensive operation in Singly linked list.
So Quick sort is well suited for arrays and not linked list.

Merge sort is a divide and conquer algorithm in which need of Random access to elements is less.
So Merge Sort can be used for sorting Linked list.


How Merge Sort works

1. Merge Sort works by breaking the linked list(or Array) into 2 equal parts say Left half and Right half.
2. Again break 2 list that we got in Step 1 in two equal parts each.  
3. Repeat above steps until only 1 element remains in Linked list (or Array) because list with only 1 element is always sorted. 
4. So in each step we are breaking the list in Left half and Right half.  
5. When complete list is divided and contains only Single element in Left and Right half each, Start comparing and sorting each Left and Right half, So that portion of Linked list will be sorted.
6. Repeat Step 5 for all the remaining Left and Right half and complete linked list will be sorted.

It works on divide and conquer technique. Time complexity is O(N log N).

Lets understand with the help of below example.

 

Sorting Linked list using Merge sort in Java. (Recursive Approach)

package linkedlist.singly;
 
public class SortLinkedList {
 
 Node startNode;
  
 public static void main(String[] args) {
  new SortLinkedList();
 }
 
 public SortLinkedList() {
  Node node1 = new Node(10);
  Node node2 = new Node(1);
  Node node3 = new Node(-2);
  Node node4 = new Node(8);
  Node node5 = new Node(9);
  Node node6 = new Node(10);
  Node node7 = new Node(1);
 
  node1.setNext(node2);
  node2.setNext(node3);
  node3.setNext(node4);
  node4.setNext(node5);
  node5.setNext(node6);
  node6.setNext(node7);
 
  startNode = node1;
   
  Node sortedStartNode = mergeSortLinkList(startNode);
  printLinkList(sortedStartNode);
 }
 
 private Node mergeSortLinkList(Node startNode){
   
  //Break the list until list is null or only 1 element is present in List.
  if(startNode==null || startNode.getNext()==null){
   return startNode;
  }
 
  //Break the linklist into 2 list.
  //Finding Middle node and then breaking the Linled list in 2 parts.
  //Now 2 list are, 1st list from start to middle and 2nd list from middle+1 to last.
   
  Node middle = getMiddle(startNode);
  Node nextOfMiddle = middle.getNext();
  middle.setNext(null);
 
  //Again breaking the List until there is only 1 element in each list.
  Node left = mergeSortLinkList(startNode);
  Node right = mergeSortLinkList(nextOfMiddle);
 
  //Once complete list is divided and contains only single element, 
  //Start merging left and right half by sorting them and passing Sorted list further. 
  Node sortedList = mergeTwoListRecursive(left, right);
   
  return sortedList;
 }
 
 //Recursive Approach for Merging Two Sorted List
 private Node mergeTwoListRecursive(Node leftStart, Node rightStart){
  if(leftStart==null)
   return rightStart;
   
  if(rightStart==null)
   return leftStart;
 
  Node temp=null;
   
  if(leftStart.getData() < rightStart.getData()){
   temp=leftStart;
   temp.setNext(mergeTwoListRecursive(leftStart.getNext(), rightStart));
  }else{
   temp=rightStart;
   temp.setNext(mergeTwoListRecursive(leftStart, rightStart.getNext()));
  }
  return temp;
 }
 
 private Node getMiddle(Node startNode) {
  if(startNode==null){
   return startNode;
  }
 
  Node pointer1=startNode;
  Node pointer2=startNode;
   
  while(pointer2!=null && pointer2.getNext()!=null && pointer2.getNext().getNext()!=null){
   pointer1 = pointer1.getNext();
   pointer2 = pointer2.getNext().getNext();
 
  }
  return pointer1;
 }
 
 private void printLinkList(Node startNode) {
  Node temp = startNode;
  while(temp!=null){
   System.out.print(temp.getData() + " ");
   temp = temp.getNext();
  }
 }
  
}

package linkedlist.singly;
 
public class SortLinkedList {
 
 Node startNode;
 
 public static void main(String[] args) {
  new SortLinkedList();
 }
 
 public SortLinkedList() {
  Node node1 = new Node(10);
  Node node2 = new Node(1);
  Node node3 = new Node(-2);
  Node node4 = new Node(8);
  Node node5 = new Node(9);
  Node node6 = new Node(10);
  Node node7 = new Node(1);
 
  node1.setNext(node2);
  node2.setNext(node3);
  node3.setNext(node4);
  node4.setNext(node5);
  node5.setNext(node6);
  node6.setNext(node7);
 
  startNode = node1;
 
  Node sortedStartNode = mergeSortLinkList(startNode);
  printLinkList(sortedStartNode);
 }
 
 private Node mergeSortLinkList(Node startNode){
 
  //Break the list until list is null or only 1 element is present in List.
  if(startNode==null || startNode.getNext()==null){
   return startNode;
  }
 
  //Break the linklist into 2 list.
  //Finding Middle node and then breaking the Linled list in 2 parts.
  //Now 2 list are, 1st list from start to middle and 2nd list from middle+1 to last.
 
  Node middle = getMiddle(startNode);
  Node nextOfMiddle = middle.getNext();
  middle.setNext(null);
 
  //Again breaking the List until there is only 1 element in each list.
  Node left = mergeSortLinkList(startNode);
  Node right = mergeSortLinkList(nextOfMiddle);
 
  //Once complete list is divided and contains only single element, 
  //Start merging left and right half by sorting them and passing Sorted list further. 
  Node sortedList = mergeTwoListIterative(left, right);
 
  return sortedList;
 }
 
 //Iterative Approach for Merging Two Sorted List
 private Node mergeTwoListIterative(Node leftStart, Node rightStart) {
 
  Node merged=null;
  Node temp=null;
 
  //To keep track of last element, so that we don't need to iterate for adding the element at last of 
  //list when either LeftStart or rightStart is NULL.
  Node lastAddedNode = null;
 
  while(leftStart!=null && rightStart!=null){
 
   if(leftStart.getData()>rightStart.getData()){
    temp = new Node(rightStart.getData());
    rightStart=rightStart.getNext();
 
   }else{
    temp = new Node(leftStart.getData());
    leftStart=leftStart.getNext();
   }
 
   if(merged==null){
    merged=temp;
   }else{
    lastAddedNode.setNext(temp);     
   }
   lastAddedNode=temp;
  }
 
  if(leftStart!=null){
   lastAddedNode.setNext(leftStart);
  }else{
   lastAddedNode.setNext(rightStart);
  }
   
  return merged;
 }
 
 private Node getMiddle(Node startNode) {
  if(startNode==null){
   return startNode;
  }
 
  Node pointer1=startNode;
  Node pointer2=startNode;
 
  while(pointer2!=null && pointer2.getNext()!=null && pointer2.getNext().getNext()!=null){
   pointer1 = pointer1.getNext();
   pointer2 = pointer2.getNext().getNext();
 
  }
  return pointer1;
 }
 
 private void printLinkList(Node startNode) {
  Node temp = startNode;
  while(temp!=null){
   System.out.print(temp.getData() + " ");
   temp = temp.getNext();
  }
 }
 
}

Given a array of integers, Sort it using Insertion sort.
Lets understand what is the input and the expected output.

Example
Assume 5 persons of different heights are visiting at your office each at 5 minutes interval and you need to make them stand according to their heights(smaller to higher). How you will do?

Below is the sequence in which person is going to visit you,

You need to arrange above persons based on their height.
First "A" will visit and that is the only person present, so just tell him to stand and you don't need to take any decision.

Second "B" will visit, so now you have to compare "B" height with "A" height, which is smaller, so shift "A" one step back and move "B" one step ahead.
Now, you have arranged 2 persons properly.

Now "C" entered in office, so you have to find proper position for "C" to stand.
First you will compare "C" height with "A", which is smaller so "A" need to be shifted one step back and "C" need to be shifted one step in front.
Now, compare "C" height with "B", which is higher than "B", So stop here because you have found proper position of "C" to stand.
Now, you have arranged 3 persons properly.

Now "E" entered in office, so you have to find proper position for "E" to stand.
First you will compare "E" height with "D", which is smaller so "D" need to be shifted one step back and "E" need to be shifted one step in front.
Now compare "E" height with "A", which is smaller so "A" need to be shifted one step back and "E" need to be shifted one step in front.
Now compare "E" height with "C", which is smaller so "C" need to be shifted one step back and "E" need to be shifted one step in front.
Now, compare "E" height with "B", which is higher than "B", So stop here because we have found proper position of "E" to stand and all the person ahead of B will be smaller than "B", so no need to check further.

Same logic we will use in Insertion sort.
Algorithm
Insertion Sort is very basic and easy sorting algorithm to understand and implement.

Insertion sort works by shifting and putting element at its correct position in array.

 Lets see how it works:

Insertion sort start's sorting an array by picking 1st element of array.
(We begin by assuming that a array with one item (position 0) is already sorted.)
So now we have one sorted elements set.

Now, we will pick 2nd element of array and compare it with sorted elements set one by one until we find correct position of 2nd element in sorted set.
Elements which are higher than 2nd element are shifted one step back to make space for 2nd element.
After putting 2nd element at its correct position, we have 2 elements in our sorted set.

Now, we will pick 3rd element of array and compare it with sorted elements set one by one until we find correct position of 3rd element in sorted set.
Elements which are higher than 3rd element are shifted one step back to make space for 3rd element.
After putting 3rd element at its correct position, we have 3 elements in our sorted set.

Repeat the process until all elements are sorted. 

Complexity
1. The insertion sort, unlike the other sorts, passes through the array only once. 
2. The insertion sort splits an array into two sub-arrays,
    First sub-array on left side is always sorted and increases in size as the sort continues.
    Second sub-array is unsorted, contains all the elements yet to be inserted into the first sub-array,
    and decreases in size as the sort continues.

3. Insertion sort is efficient for (quite) small data sets.
4. It is more efficient than selection sort or bubble sort.
5. Insertion sort is very efficient for arrays which is nearly(almost) sorted and it sorts nearly sorted
    array in time complexity of O(N).
    (For sorting an array containing elements in descending order to ascending order, insertion sort
    will give poor performance and complexity will be O(N^2))

6. It is Stable sort; i.e., does not change the relative order of elements with equal keys.
7. It is In-place sort; i.e., only requires a constant amount O(1) of additional memory space
8. It can sort elements as it receives it and no need of complete data initially before start sorting.
    (Online).
Java Program for Insertion Sort

package sorting;
 
public class InsertionSort {
 public static void main(String[] args) {
  new InsertionSort();
 }
 
 public InsertionSort() {
  int[] arr=new int[]{1,9,4,10,0};
 
  System.out.println("Before Sorting...");
  printArray(arr);
   
  insertionSort(arr);
   
  System.out.println("\nAfter Sorting...");
  printArray(arr);
 }
 
 private void insertionSort(int arr[]){
  if(arr==null || arr.length<2){
   return;
  }
 
  for (int i = 1; i < arr.length; i++) {
   int temp = arr[i];
    
   // Comparison starts from one step back of element on which we are working that is i.
   int j=i-1; 
    
   //Compare elements till we not found element higher than temp or all element are compared.
   while( j >= 0 && arr[j] > temp){
    arr[j+1] = arr[j];
    j--;
   }
   arr[j+1]=temp;   
  }
 }
 
 private void printArray(int arr[]){
  for (int i = 0; i < arr.length; i++) {
   System.out.print(arr[i] + " ");
  }
 }
  
}