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Find Sum of Even and Odd Nodes in a Linked List in C++
In this problem, we are given a linked list. Our task is to find the sum of even and odd nodes in a linked list.
Let's take an example to understand the problem,
Input : linked list : 3 -> 2 -> 5 -> 7 -> 1 -> 9 Output : evenSum = 2 ; oddSum = 25
Explanation −
evenSum = 2 oddSum = 3 + 5 + 7 + 1 + 9 = 25
Solution Approach
A simple approach to solve the problem is traversing the linked list and checking for even or odd values and adding them to their respective sum value.
Algorithm
-
Step 1 − Traverse the linked list.
Step 1.1 − If the value of the current node is even, add it to evenSum.
Step 1.2 − If the value of the current node is odd, add it to oddSum.
Step 2 − Return oddSum and evenSum.
Example
Program to illustrate the working of our solution
#include <iostream> using namespace std; struct Node { int data; Node* next; }; void insertNode(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; } } bool isEven(int a){ return (a % 2); } void findEvenAndOddSum(Node* root) { int oddSum = 0, evenSum = 0; Node* node = root; while (node != NULL) { if (isEven(node->data)) evenSum += node->data; else oddSum += node->data; node = node->next; } cout<<"Sum of nodes with even value is "<<evenSum<<endl; cout<<"Sum of nodes with odd value is "<<oddSum; } int main() { Node* root = NULL; insertNode(&root, 3); insertNode(&root, 2); insertNode(&root, 5); insertNode(&root, 7); insertNode(&root, 1); insertNode(&root, 9); insertNode(&root, 6); findEvenAndOddSum(root); return 0; }
Output
Sum of nodes with even value is 25 Sum of nodes with odd value is 8
Updated on: 2022-01-27T08:18:57+05:30
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