Finding an Element in a LinkedList - C++ - c++

Question

I am trying to write a function find(Node* node, int valueInput) that starts at the given node and returns the index of the node with the value valueInput, and return -1 if valueInput does not exist.
Here is the code
#include <iostream>
class Node {
public:
int value;
Node* next = NULL;
};
int find(Node* node, int valueInput)
{
Node* tempNode = node;
int count = 0;
while (tempNode != NULL)
{
if (tempNode->value == 0)
{
break;
}
++count;
tempNode = tempNode->next;
}
return -1;
}
int main()
{
int valueInput, currentValue;
int listValues;
char arrow;
Node* node1, * previous, * head;
std::cin >> valueInput;
std::cin >> currentValue;
node1 = new Node();
node1->value = currentValue;
node1->next = NULL;
previous = node1;
head = node1;
while (std::cin)
{
std::cin >> arrow;
std::cin >> arrow;
std::cin >> currentValue;
node1 = new Node();
node1->value = currentValue;
previous->next = node1;
previous = node1;
}
std::cout << find(head, valueInput);
}
Currently, my program returns -1 always
Sample input and outputs:
Sample Input:
5
1->2->5
Sample Output:
2

Answer 1

That's because your code only has a return -1 statement. You should return count instead of break. Also, you are always comparing against 0 instead of the valueInput
int find(Node* node, int valueInput)
{
Node* tempNode = node;
int count = 0;
while (tempNode != nullptr)
{
if (tempNode->value == valueInput)
{
return count;
}
++count;
tempNode = tempNode->next;
}
return -1;
}
You should also be using nullptr instead of NULL