# Maximum sum contiguous nodes in the given linked list

Given a linked list, the task is to find the maximum sum for any contiguous nodes.

Examples:

Input: -2 -> -3 -> 4 -> -1 -> -2 -> 1 -> 5 -> -3 -> NULL
Output: 7
4 -> -1 -> -2 -> 1 -> 5 is the sub-list with the given sum.

Input: 1 -> 2 -> 3 -> 4 -> NULL
Output: 10

## Recommended: Please try your approach on {IDE} first, before moving on to the solution.

Approach: Kadane’s algorithm has been discussed in this article to work on arrays to find the maximum sub-array sum but it can be modified to work on linked lists too. Since Kadane’s algorithm doesn’t require to access random elements, it is also applicable on the linked lists in linear time.

Below is the implementation of the above approach:

## C++

 `// C++ implementation of the approach ` `#include ` `using` `namespace` `std; ` ` `  `// A linked list node ` `class` `Node { ` `public``: ` `    ``int` `data; ` `    ``Node* next; ` `}; ` ` `  `/* Given a reference (pointer to pointer) to the head  ` `of a list and an int, appends a new node at the end */` `void` `append(Node** head_ref, ``int` `new_data) ` `{ ` `    ``// Allocate node ` `    ``Node* new_node = ``new` `Node(); ` ` `  `    ``Node* last = *head_ref; ``/* used in step 5*/` ` `  `    ``// Put in the data ` `    ``new_node->data = new_data; ` ` `  `    ``/* This new node is going to be  ` `    ``the last node, so make next of  ` `    ``it as NULL*/` `    ``new_node->next = NULL; ` ` `  `    ``/* If the Linked List is empty,  ` `    ``then make the new node as head */` `    ``if` `(*head_ref == NULL) { ` `        ``*head_ref = new_node; ` `        ``return``; ` `    ``} ` ` `  `    ``// Else traverse till the last node ` `    ``while` `(last->next != NULL) ` `        ``last = last->next; ` ` `  `    ``// Change the next of last node ` `    ``last->next = new_node; ` `    ``return``; ` `} ` ` `  `// Function to return the maximum contiguous ` `// nodes sum in the given linked list ` `int` `MaxContiguousNodeSum(Node* head) ` `{ ` ` `  `    ``// If the list is empty ` `    ``if` `(head == NULL) ` `        ``return` `0; ` ` `  `    ``// If the list contains a single element ` `    ``if` `(head->next == NULL) ` `        ``return` `head->data; ` ` `  `    ``// max_ending_here will store the maximum sum ` `    ``// ending at the current node, currently it ` `    ``// will be initialised to the maximum sum ending ` `    ``// at the first node which is the first node's value ` `    ``int` `max_ending_here = head->data; ` ` `  `    ``// max_so_far will store the maximum sum of ` `    ``// contiguous nodes so far which is the required ` `    ``// answer at the end of the linked list traversal ` `    ``int` `max_so_far = head->data; ` ` `  `    ``// Starting from the second node ` `    ``head = head->next; ` ` `  `    ``// While there are nodes in linked list ` `    ``while` `(head != NULL) { ` ` `  `        ``// max_ending_here will be the maximum of either ` `        ``// the current node's value or the current node's ` `        ``// value added with the max_ending_here ` `        ``// for the previous node ` `        ``max_ending_here = max(head->data, ` `                              ``max_ending_here + head->data); ` ` `  `        ``// Update the maximum sum so far ` `        ``max_so_far = max(max_ending_here, max_so_far); ` ` `  `        ``// Get to the next node ` `        ``head = head->next; ` `    ``} ` ` `  `    ``// Return the maximum sum so far ` `    ``return` `max_so_far; ` `} ` ` `  `// Driver code ` `int` `main() ` `{ ` `    ``// Create the linked list ` `    ``Node* head = NULL; ` `    ``append(&head, -2); ` `    ``append(&head, -3); ` `    ``append(&head, 4); ` `    ``append(&head, -1); ` `    ``append(&head, -2); ` `    ``append(&head, 1); ` `    ``append(&head, 5); ` `    ``append(&head, -3); ` ` `  `    ``cout << MaxContiguousNodeSum(head); ` ` `  `    ``return` `0; ` `} `

## Java

 `// Java implementation of the approach ` `class` `GFG ` `{ ` ` `  `// A linked list node ` `static` `class` `Node  ` `{ ` `    ``int` `data; ` `    ``Node next; ` `}; ` ` `  `/* Given a reference (pointer to pointer) to the head  ` `of a list and an int, appends a new node at the end */` `static` `Node append(Node head_ref, ``int` `new_data) ` `{ ` `    ``// Allocate node ` `    ``Node new_node = ``new` `Node(); ` ` `  `    ``Node last = head_ref; ``/* used in step 5*/` ` `  `    ``// Put in the data ` `    ``new_node.data = new_data; ` ` `  `    ``/* This new node is going to be  ` `    ``the last node, so make next of  ` `    ``it as null*/` `    ``new_node.next = ``null``; ` ` `  `    ``/* If the Linked List is empty,  ` `    ``then make the new node as head */` `    ``if` `(head_ref == ``null``)  ` `    ``{ ` `        ``head_ref = new_node; ` `        ``return` `head_ref; ` `    ``} ` ` `  `    ``// Else traverse till the last node ` `    ``while` `(last.next != ``null``) ` `        ``last = last.next; ` ` `  `    ``// Change the next of last node ` `    ``last.next = new_node; ` `    ``return` `head_ref; ` `} ` ` `  `// Function to return the maximum contiguous ` `// nodes sum in the given linked list ` `static` `int` `MaxContiguousNodeSum(Node head) ` `{ ` ` `  `    ``// If the list is empty ` `    ``if` `(head == ``null``) ` `    ``return` `0``; ` ` `  `    ``// If the list contains a single element ` `    ``if` `(head.next == ``null``) ` `        ``return` `head.data; ` ` `  `    ``// max_ending_here will store the maximum sum ` `    ``// ending at the current node, currently it ` `    ``// will be initialised to the maximum sum ending ` `    ``// at the first node which is the first node's value ` `    ``int` `max_ending_here = head.data; ` ` `  `    ``// max_so_far will store the maximum sum of ` `    ``// contiguous nodes so far which is the required ` `    ``// answer at the end of the linked list traversal ` `    ``int` `max_so_far = head.data; ` ` `  `    ``// Starting from the second node ` `    ``head = head.next; ` ` `  `    ``// While there are nodes in linked list ` `    ``while` `(head != ``null``)  ` `    ``{ ` ` `  `        ``// max_ending_here will be the maximum of either ` `        ``// the current node's value or the current node's ` `        ``// value added with the max_ending_here ` `        ``// for the previous node ` `        ``max_ending_here = Math.max(head.data, ` `                            ``max_ending_here + head.data); ` ` `  `        ``// Update the maximum sum so far ` `        ``max_so_far = Math.max(max_ending_here, max_so_far); ` ` `  `        ``// Get to the next node ` `        ``head = head.next; ` `    ``} ` ` `  `    ``// Return the maximum sum so far ` `    ``return` `max_so_far; ` `} ` ` `  `// Driver code ` `public` `static` `void` `main(String[] args) ` `{ ` `    ``// Create the linked list ` `    ``Node head = ``null``; ` `    ``head = append(head, -``2``); ` `    ``head = append(head, -``3``); ` `    ``head = append(head, ``4``); ` `    ``head = append(head, -``1``); ` `    ``head = append(head, -``2``); ` `    ``head = append(head, ``1``); ` `    ``head = append(head, ``5``); ` `    ``head = append(head, -``3``); ` ` `  `    ``System.out.print(MaxContiguousNodeSum(head)); ` `} ` `} ` ` `  `// This code is contributed by PrinciRaj1992 `

## C#

 `// C# implementation of the approach ` `using` `System; ` `class` `GFG ` `{ ` ` `  `// A linked list node ` `public` `class` `Node  ` `{ ` `    ``public` `int` `data; ` `    ``public` `Node next; ` `}; ` ` `  `/* Given a reference (pointer to pointer) to the head  ` `of a list and an int, appends a new node at the end */` `static` `Node append(Node head_ref, ``int` `new_data) ` `{ ` `    ``// Allocate node ` `    ``Node new_node = ``new` `Node(); ` ` `  `    ``Node last = head_ref; ``/* used in step 5*/` ` `  `    ``// Put in the data ` `    ``new_node.data = new_data; ` ` `  `    ``/* This new node is going to be  ` `    ``the last node, so make next of  ` `    ``it as null*/` `    ``new_node.next = ``null``; ` ` `  `    ``/* If the Linked List is empty,  ` `    ``then make the new node as head */` `    ``if` `(head_ref == ``null``)  ` `    ``{ ` `        ``head_ref = new_node; ` `        ``return` `head_ref; ` `    ``} ` ` `  `    ``// Else traverse till the last node ` `    ``while` `(last.next != ``null``) ` `        ``last = last.next; ` ` `  `    ``// Change the next of last node ` `    ``last.next = new_node; ` `    ``return` `head_ref; ` `} ` ` `  `// Function to return the maximum contiguous ` `// nodes sum in the given linked list ` `static` `int` `MaxContiguousNodeSum(Node head) ` `{ ` ` `  `    ``// If the list is empty ` `    ``if` `(head == ``null``) ` `    ``return` `0; ` ` `  `    ``// If the list contains a single element ` `    ``if` `(head.next == ``null``) ` `        ``return` `head.data; ` ` `  `    ``// max_ending_here will store the maximum sum ` `    ``// ending at the current node, currently it ` `    ``// will be initialised to the maximum sum ending ` `    ``// at the first node which is the first node's value ` `    ``int` `max_ending_here = head.data; ` ` `  `    ``// max_so_far will store the maximum sum of ` `    ``// contiguous nodes so far which is the required ` `    ``// answer at the end of the linked list traversal ` `    ``int` `max_so_far = head.data; ` ` `  `    ``// Starting from the second node ` `    ``head = head.next; ` ` `  `    ``// While there are nodes in linked list ` `    ``while` `(head != ``null``)  ` `    ``{ ` ` `  `        ``// max_ending_here will be the maximum of either ` `        ``// the current node's value or the current node's ` `        ``// value added with the max_ending_here ` `        ``// for the previous node ` `        ``max_ending_here = Math.Max(head.data, ` `                                   ``max_ending_here +  ` `                                   ``head.data); ` ` `  `        ``// Update the maximum sum so far ` `        ``max_so_far = Math.Max(max_ending_here,  ` `                              ``max_so_far); ` ` `  `        ``// Get to the next node ` `        ``head = head.next; ` `    ``} ` ` `  `    ``// Return the maximum sum so far ` `    ``return` `max_so_far; ` `} ` ` `  `// Driver code ` `public` `static` `void` `Main(String[] args) ` `{ ` `    ``// Create the linked list ` `    ``Node head = ``null``; ` `    ``head = append(head, -2); ` `    ``head = append(head, -3); ` `    ``head = append(head, 4); ` `    ``head = append(head, -1); ` `    ``head = append(head, -2); ` `    ``head = append(head, 1); ` `    ``head = append(head, 5); ` `    ``head = append(head, -3); ` ` `  `    ``Console.Write(MaxContiguousNodeSum(head)); ` `} ` `} ` ` `  `// This code is contributed by 29AjayKumar `

Output:

```7
```

My Personal Notes arrow_drop_up Check out this Author's contributed articles.

If you like GeeksforGeeks and would like to contribute, you can also write an article using contribute.geeksforgeeks.org or mail your article to contribute@geeksforgeeks.org. See your article appearing on the GeeksforGeeks main page and help other Geeks.

Please Improve this article if you find anything incorrect by clicking on the "Improve Article" button below.

Improved By : princiraj1992, 29AjayKumar

Article Tags :
Practice Tags :

Be the First to upvote.

Please write to us at contribute@geeksforgeeks.org to report any issue with the above content.