Maximum sum contiguous nodes in the given linked list
Last Updated :
12 Jul, 2025
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
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 <bits/stdc++.h>
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 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
# Python3 implementation of the approach
# A linked list node
class Node:
def __init__(self, data):
self.data = data
self.next = None
# Given a reference (pointer to pointer)
# to the head of a list and an int,
# appends a new node at the end
def append(head_ref, new_data):
# Allocate node
new_node = Node(new_data)
last = head_ref # used in step 5
# This new node is going to be
# the last node, so make next of
# it as None
new_node.next = None
# If the Linked List is empty,
# then make the new node as head
if (head_ref == None):
head_ref = new_node
return head_ref
# Else traverse till the last node
while (last.next != None):
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
def MaxContiguousNodeSum(head):
# If the list is empty
if (head == None):
return 0
# If the list contains a single element
if (head.next == None):
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
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
max_so_far = head.data
# Starting from the second node
head = head.next
# While there are nodes in linked list
while (head != None):
# 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
if __name__=='__main__':
# Create the linked list
head = None
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)
print(MaxContiguousNodeSum(head))
# This code is contributed by rutvik_56
// 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
<script>
// JavaScript implementation of the approach
// Structure of a node of the linked list
class Node {
constructor() {
this.data = 0;
this.next = null;
}
}
/* Given a reference (pointer to pointer) to the head
of a list and an int, appends a new node at the end */
function append( head_ref, new_data)
{
// Allocate node
var new_node = new Node();
var 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
function MaxContiguousNodeSum( 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
let 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
let 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
// Create the linked list
var 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);
document.write(MaxContiguousNodeSum(head));
</script>
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