# 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 the original head is not the smallest value in the linked list.

```MergeSort(headRef)
1) If the 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)
```

## C++

 `// C++ code for linked list merged sort ` `#include ` `using` `namespace` `std; ` ` `  `/* Link list node */` `class` `Node { ` `public``: ` `    ``int` `data; ` `    ``Node* next; ` `}; ` ` `  `/* function prototypes */` `Node* SortedMerge(Node* a, Node* b); ` `void` `FrontBackSplit(Node* source, ` `                    ``Node** frontRef, Node** backRef); ` ` `  `/* sorts the linked list by changing next pointers (not data) */` `void` `MergeSort(Node** headRef) ` `{ ` `    ``Node* head = *headRef; ` `    ``Node* a; ` `    ``Node* b; ` ` `  `    ``/* Base case -- length 0 or 1 */` `    ``if` `((head == NULL) || (head->next == NULL)) { ` `        ``return``; ` `    ``} ` ` `  `    ``/* Split head into 'a' and 'b' sublists */` `    ``FrontBackSplit(head, &a, &b); ` ` `  `    ``/* Recursively sort the sublists */` `    ``MergeSort(&a); ` `    ``MergeSort(&b); ` ` `  `    ``/* answer = merge the two sorted lists together */` `    ``*headRef = SortedMerge(a, b); ` `} ` ` `  `/* See https:// www.geeksforgeeks.org/?p=3622 for details of this  ` `function */` `Node* SortedMerge(Node* a, Node* b) ` `{ ` `    ``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(Node* source, ` `                    ``Node** frontRef, Node** backRef) ` `{ ` `    ``Node* fast; ` `    ``Node* slow; ` `    ``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(Node* node) ` `{ ` `    ``while` `(node != NULL) { ` `        ``cout << node->data << ``" "``; ` `        ``node = node->next; ` `    ``} ` `} ` ` `  `/* Function to insert a node at the beginging of the linked list */` `void` `push(Node** head_ref, ``int` `new_data) ` `{ ` `    ``/* allocate node */` `    ``Node* new_node = ``new` `Node(); ` ` `  `    ``/* put in the data */` `    ``new_node->data = new_data; ` ` `  `    ``/* link the old list off the new node */` `    ``new_node->next = (*head_ref); ` ` `  `    ``/* move the head to point to the new node */` `    ``(*head_ref) = new_node; ` `} ` ` `  `/* Drier program to test above functions*/` `int` `main() ` `{ ` `    ``/* Start with the empty list */` `    ``Node* res = NULL; ` `    ``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); ` ` `  `    ``cout << ``"Sorted Linked List is: \n"``; ` `    ``printList(a); ` ` `  `    ``return` `0; ` `} ` ` `  `// This is code is contributed by rathbhupendra `

## C

 `// C code for linked list merged sort ` `#include ` `#include ` ` `  `/* 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* head = *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 */` `    ``FrontBackSplit(head, &a, &b); ` ` `  `    ``/* Recursively sort the sublists */` `    ``MergeSort(&a); ` `    ``MergeSort(&b); ` ` `  `    ``/* answer = merge the two sorted lists together */` `    ``*headRef = SortedMerge(a, b); ` `} ` ` `  `/* See https:// 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; ` `    ``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 */` `    ``new_node->next = (*head_ref); ` ` `  `    ``/* move the head to point to the new node */` `    ``(*head_ref) = 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``(``"Sorted Linked List is: \n"``); ` `    ``printList(a); ` ` `  `    ``getchar``(); ` `    ``return` `0; ` `} `

## Java

 `// Java program to illustrate merge sorted ` `// of linkedList ` ` `  `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 ` `    ``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 */` `        ``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(); ` `        ``/* ` `         ``* Let us create a unsorted linked list to test the functions ` `         ``* created. The list 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``); ` ` `  `        ``// Apply merge Sort ` `        ``li.head = li.mergeSort(li.head); ` `        ``System.out.print(``"\n Sorted Linked List is: \n"``); ` `        ``li.printList(li.head); ` `    ``} ` `} ` ` `  `// This code is contributed by Rishabh Mahrsee `

## C#

 `// C# program to illustrate merge sorted  ` `// of linkedList  ` `using` `System; ` ` `  `public` `class` `linkedList ` `{  ` `    ``node head = ``null``;  ` `     `  `    ``// node a, b;  ` `    ``public` `class` `node  ` `    ``{  ` `        ``public` `int` `val;  ` `        ``public` `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 */` `        ``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``)  ` `        ``{  ` `            ``Console.Write(headref.val + ``" "``);  ` `            ``headref = headref.next;  ` `        ``}  ` `    ``}  ` ` `  `    ``// Driver code ` `    ``public` `static` `void` `Main(String []args)  ` `    ``{  ` ` `  `        ``linkedList li = ``new` `linkedList();  ` `        ``/*  ` `        ``* Let us create a unsorted linked list to test the functions  ` `        ``* created. The list 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);  ` ` `  `        ``// Apply merge Sort  ` `        ``li.head = li.mergeSort(li.head);  ` `        ``Console.Write(``"\n Sorted Linked List is: \n"``);  ` `        ``li.printList(li.head);  ` `    ``}  ` `}  ` ` `  `// This code is contributed by Arnab Kundu `

Output:

```Sorted Linked List is:
2 3 5 10 15 20
```

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.

My Personal Notes arrow_drop_up

Improved By : rathbhupendra, andrew1234

Article Tags :

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