Find sum of even and odd nodes in a linked list

Last Updated : 12 Sep, 2022

Given a linked list, the task is to find the sum of even and odd nodes in it separately.

Examples:

Input: 1 -> 2 -> 3 -> 4 -> 5 -> 6 -> 7
Output:
Even Sum = 12
Odd Sum = 16

Input: 5 -> 7 -> 8 -> 10 -> 15
Output:
Even Sum = 18
Odd Sum = 27

Approach: Traverse the whole linked list and for each node:-

If the element is even then we add that element to the variable which is holding the sum of even elements.
If the element is odd then we add that element to the variable which is holding the sum of odd elements.

Below is the implementation of the above approach:

C++

// C++ implementation of the approach
#include <iostream>
using namespace std;

// Represents node of the linked list
struct Node {
    int data;
    Node* next;
};

// Function to insert a node at the
// end of the linked list
void insert(Node** root, int item)
{
    Node *ptr = *root, *temp = new Node;
    temp->data = item;
    temp->next = NULL;

    if (*root == NULL)
        *root = temp;
    else {
        while (ptr->next != NULL)
            ptr = ptr->next;
        ptr->next = temp;
    }
}

// Function to print the sum of even
// and odd nodes of the linked lists
void evenOdd(Node* root)
{
    int odd = 0, even = 0;
    Node* ptr = root;
    while (ptr != NULL) {

        // If current node's data is even
        if (ptr->data % 2 == 0)
            even += ptr->data;

        // If current node's data is odd
        else
            odd += ptr->data;

        // ptr now points to the next node
        ptr = ptr->next;
    }

    cout << "Even Sum = " << even << endl;
    cout << "Odd Sum = " << odd << endl;
}

// Driver code
int main()
{
    Node* root = NULL;
    insert(&root, 1);
    insert(&root, 2);
    insert(&root, 3);
    insert(&root, 4);
    insert(&root, 5);
    insert(&root, 6);
    insert(&root, 7);

    evenOdd(root);

    return 0;
}

Java

// Java implementation of the approach 
class GfG 
{

// Represents node of the linked list 
static class Node 
{ 
    int data; 
    Node next; 
}
static Node root;

// Function to insert a node at the 
// end of the linked list 
static void insert(int item) 
{ 
    Node ptr = root, temp = new Node(); 
    temp.data = item; 
    temp.next = null; 

    if (root == null) 
        root = temp; 
    else 
    { 
        while (ptr.next != null) 
            ptr = ptr.next; 
        ptr.next = temp; 
    } 
} 

// Function to print the sum of even 
// and odd nodes of the linked lists 
static void evenOdd(Node root) 
{ 
    int odd = 0, even = 0; 
    Node ptr = root; 
    while (ptr != null) 
    { 

        // If current node's data is even 
        if (ptr.data % 2 == 0) 
            even += ptr.data; 

        // If current node's data is odd 
        else
            odd += ptr.data; 

        // ptr now points to the next node 
        ptr = ptr.next; 
    } 

    System.out.println("Even Sum = " + even); 
    System.out.println("Odd Sum = " + odd); 
} 

// Driver code 
public static void main(String[] args) 
{ 
    // Node* root = NULL; 
    insert( 1); 
    insert( 2); 
    insert( 3); 
    insert( 4); 
    insert(5); 
    insert(6); 
    insert( 7); 

    evenOdd(root); 
}
} 

// This code is contributed by Prerna Saini

Python3

# Python3 implementation of the approach
import math

# Represents node of the linked list
class Node: 
    def __init__(self, data): 
        self.data = data 
        self.next = None

# Function to insert a node at the
# end of the linked list
def insert(root, item):
    ptr = root
    temp = Node(item)
    temp.data = item
    temp.next = None

    if (root == None):
        root = temp
    else:
        while (ptr.next != None):
            ptr = ptr.next
        ptr.next = temp
    
    return root

# Function to print the sum of even
# and odd nodes of the linked lists
def evenOdd(root):
    odd = 0
    even = 0
    ptr = root
    while (ptr != None):

        # If current node's data is even
        if (ptr.data % 2 == 0):
            even = even + ptr.data

        # If current node's data is odd
        else:
            odd = odd + ptr.data

        # ptr now points to the next node
        ptr = ptr.next
    
    print( "Even Sum = ", even)
    print( "Odd Sum = ", odd)

# Driver code
if __name__=='__main__': 
    root = None
    root = insert(root, 1)
    root = insert(root, 2)
    root = insert(root, 3)
    root = insert(root, 4)
    root = insert(root, 5)
    root = insert(root, 6)
    root = insert(root, 7)

    evenOdd(root)

# This code is contributed by AbhiThakur

// C# implementation of the approach 
using System;

class GfG 
{ 

// Represents node of the linked list 
public class Node 
{ 
    public int data; 
    public Node next; 
} 
static Node root; 

// Function to insert a node at the 
// end of the linked list 
static void insert(int item) 
{ 
    Node ptr = root, temp = new Node(); 
    temp.data = item; 
    temp.next = null; 

    if (root == null) 
        root = temp; 
    else
    { 
        while (ptr.next != null) 
            ptr = ptr.next; 
        ptr.next = temp; 
    } 
} 

// Function to print the sum of even 
// and odd nodes of the linked lists 
static void evenOdd(Node root) 
{ 
    int odd = 0, even = 0; 
    Node ptr = root; 
    while (ptr != null) 
    { 

        // If current node's data is even 
        if (ptr.data % 2 == 0) 
            even += ptr.data; 

        // If current node's data is odd 
        else
            odd += ptr.data; 

        // ptr now points to the next node 
        ptr = ptr.next; 
    } 

    Console.WriteLine("Even Sum = " + even); 
    Console.WriteLine("Odd Sum = " + odd); 
} 

// Driver code 
public static void Main(String []args) 
{ 
    // Node* root = NULL; 
    insert( 1); 
    insert( 2); 
    insert( 3); 
    insert( 4); 
    insert(5); 
    insert(6); 
    insert( 7); 

    evenOdd(root); 
} 
} 

// This code is contributed by Arnab Kundu

JavaScript

<script>

// Javascript implementation of the approach

// Represents node of the linked list
class Node {
        constructor() {
                this.data = 0;
                this.next = null;
             }
        }
        
        
// Function to insert a node at the
// end of the linked list
function insert( item)
{
    var ptr = root, temp = new Node();
    temp.data = item;
    temp.next = null;

    if (root == null)
        root = temp;
    else
    {
        while (ptr.next != null)
            ptr = ptr.next;
        ptr.next = temp;
    }
}

// Function to print the sum of even
// and odd nodes of the linked lists
function evenOdd( root)
{
    let odd = 0, even = 0;
    let ptr = root;
    while (ptr != null)
    {

        // If current node's data is even
        if (ptr.data % 2 == 0)
            even += ptr.data;

        // If current node's data is odd
        else
            odd += ptr.data;

        // ptr now points to the next node
        ptr = ptr.next;
    }

    document.write("Even Sum = " + even);
    document.write("</br>");
    document.write("Odd Sum = " + odd);
}


// Driver Code

var root = null;
insert( 1);
insert( 2);
insert( 3);
insert( 4);
insert(5);
insert(6);
insert( 7);

evenOdd(root);

// This code is contributed by jana_sayantan.
</script>

Output:

Even Sum = 12
Odd Sum = 16

Time complexity: O(N) where N is number of nodes in the given linked list.
Auxiliary space: O(1), as constant space is used.

Find the balanced node in a Linked List

Premdeep Toppo

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