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 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 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
<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.
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