C++ Ordered Linked List Search Function Algorithm logic

Question

I'm new to C++ and I'm trying to write an algorithm to search a linked list, but I'm having a little trouble with my logic. The ??? question marks in bold are the parts I'm having trouble with. I appreciate any help with this.

  ListNode *MyLinkedList::Search(int key)
{
    ListNode *temp = head;  // Assume ListNode is a structure and contains the variable int key;
    // Search for the key
    while((temp != NULL) && (key != temp->key))
    {
        temp = temp -> next; // Advance to next node
    {
    if(**???**)     // Make sure node is found
    {   
        return **???**; // If found, return appropriate value
    }
    else 
    {   
        return NULL;  // return NULL when not found
    }
}

Answer 1

If the key was found key == temp->key will be true and temp != NULL will be false, so:

if(key == temp->key)     // Make sure node is found
{   
    return temp; // If found, return appropriate value
}

OR:

if (temp != NULL)     // Make sure node is found
{   
    return temp; // If found, return appropriate value
}

Answer 2

Try this code:

ListNode *MyLinkedList::Search(int key, ListNode *head)
{
    ListNode *temp = head;  // Assume ListNode is a structure and contains the variable         int key;
    // Search for the key
    while(temp != NULL)
    {
        if (key == temp->key)
           return temp;
        temp = temp -> next; // Advance to next node
    }
    return NULL;  // return NULL when not found
}

EDIT

You should use the list from stl and the find algorithm, if you aren't required to write your own container; They are tested and safe:

http://en.cppreference.com/w/cpp/container/list

http://en.cppreference.com/w/cpp/algorithm/find

Answer 3

You don't need an if . Just return temp . If the correct key is present in the list, temp would be pointing to it, otherwise it's NULL .

Answer 4

This will work for you

if(temp != NULL) // Make sure node is found
{
    return temp; // Make sure node is found
}

Answer 5

You could do this:

if(temp != NULL)     // Make sure node is found
{   
    return temp; // If found, return appropriate value
}

But much simpler is just

return temp;

since if temp is null, you want to return null anyway.