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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: 27-Jan-2022
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