# 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.

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.

```MergeSort(headRef)
1) If head is NULL or there is only one element in the Linked List
then return.
2) Else divide the linked list into two halves.
FrontBackSplit(head, &a, &b); /* a and b are two halves */
3) Sort the two halves a and b.
MergeSort(a);
MergeSort(b);
4) Merge the sorted a and b (using SortedMerge() discussed here)
and update the head pointer using headRef.
*headRef = SortedMerge(a, b);
```

## C

```#include<stdio.h>
#include<stdlib.h>

/* Link list node */
struct Node
{
int data;
struct Node* next;
};

/* function prototypes */
struct Node* SortedMerge(struct Node* a, struct Node* b);
void FrontBackSplit(struct Node* source,
struct Node** frontRef, struct Node** backRef);

/* sorts the linked list by changing next pointers (not data) */
void MergeSort(struct Node** headRef)
{
struct Node* a;
struct Node* b;

/* Base case -- length 0 or 1 */
if ((head == NULL) || (head->next == NULL))
{
return;
}

/* Split head into 'a' and 'b' sublists */

/* Recursively sort the sublists */
MergeSort(&a);
MergeSort(&b);

/* answer = merge the two sorted lists together */
*headRef = SortedMerge(a, b);
}

/* See http://www.geeksforgeeks.org/?p=3622 for details of this
function */
struct Node* SortedMerge(struct Node* a, struct Node* b)
{
struct Node* result = NULL;

/* Base cases */
if (a == NULL)
return(b);
else if (b==NULL)
return(a);

/* Pick either a or b, and recur */
if (a->data <= b->data)
{
result = a;
result->next = SortedMerge(a->next, b);
}
else
{
result = b;
result->next = SortedMerge(a, b->next);
}
return(result);
}

/* UTILITY FUNCTIONS */
/* Split the nodes of the given list into front and back halves,
and return the two lists using the reference parameters.
If the length is odd, the extra node should go in the front list.
Uses the fast/slow pointer strategy.  */
void FrontBackSplit(struct Node* source,
struct Node** frontRef, struct Node** backRef)
{
struct Node* fast;
struct Node* slow;
if (source==NULL || source->next==NULL)
{
/* length < 2 cases */
*frontRef = source;
*backRef = NULL;
}
else
{
slow = source;
fast = source->next;

/* Advance 'fast' two nodes, and advance 'slow' one node */
while (fast != NULL)
{
fast = fast->next;
if (fast != NULL)
{
slow = slow->next;
fast = fast->next;
}
}

/* 'slow' is before the midpoint in the list, so split it in two
at that point. */
*frontRef = source;
*backRef = slow->next;
slow->next = NULL;
}
}

/* Function to print nodes in a given linked list */
void printList(struct Node *node)
{
while(node!=NULL)
{
printf("%d ", node->data);
node = node->next;
}
}

/* Function to insert a node at the beginging of the linked list */
void push(struct Node** head_ref, int new_data)
{
/* allocate node */
struct Node* new_node =
(struct Node*) malloc(sizeof(struct Node));

/* put in the data  */
new_node->data  = new_data;

/* link the old list off the new node */

/* move the head to point to the new node */
}

/* Drier program to test above functions*/
int main()
{
/* Start with the empty list */
struct Node* res = NULL;
struct Node* a = NULL;

/* Let us create a unsorted linked lists to test the functions
Created lists shall be a: 2->3->20->5->10->15 */
push(&a, 15);
push(&a, 10);
push(&a, 5);
push(&a, 20);
push(&a, 3);
push(&a, 2);

/* Sort the above created Linked List */
MergeSort(&a);

printf("\n Sorted Linked List is: \n");
printList(a);

getchar();
return 0;
}
```

## Java

```// Java program to illustrate merge sorted

{
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
node getMiddle(node h)
{
//Base case
if (h == null)
return h;
node fastptr = h.next;
node slowptr = h;

// Move fastptr by two and slow ptr by one
// Finally slowptr will point to middle node
while (fastptr != null)
{
fastptr = fastptr.next;
if(fastptr!=null)
{
slowptr = slowptr.next;
fastptr=fastptr.next;
}
}
return slowptr;
}

void push(int new_data)
{
/* allocate node */
node new_node = new node(new_data);

/* link the old list off the new node */

/* move the head to point to the new node */
}

// Utility function to print the linked list
{
while (headref != null)
{
System.out.print(headref.val + " ");
}
}

public static void main(String[] args)
{

/*
* Let us create a unsorted linked lists to test the functions Created
* lists shall be a: 2->3->20->5->10->15
*/
li.push(15);
li.push(10);
li.push(5);
li.push(20);
li.push(3);
li.push(2);
System.out.println("Linked List without sorting is :");

// Apply merge Sort
System.out.print("\n Sorted Linked List is: \n");
}
}

// This code is contributed by Rishabh Mahrsee
```

Time Complexity: O(n Log n)

Please write comments if you find the above code/algorithm incorrect, or find better ways to solve the same problem.

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