Linked list – recursive functions

A linked list, is basically a list of elements that have a link attached to each one to the next. So for example if you have a structure like

struct List {
     string name;
     List *next;
};

then the List has a internals of a name which is of type string, and then a link (a pointer) to the next List in the list of linked List types. I have called it a Elem in the source code below because then it may make more sense, since it is kinder a element of the whole linked list.

So to create a I wrote a function to add a element (Elem) to the list already, since we are going to be altering the list then need to pass as a reference (&) so we actually alter the main variable and not just the functions variable and it is not passed back.

struct Elem {
	string name;
	Elem *next;
};
 
// inserts a Elem into a linked list
void setUpElem(Elem * & list, Elem *addnew)
{
	addnew->next = list;
	list = addnew;
}
 
.. 
int main()
{
    Elem *elem = NULL;
    Elem newElem;
    newElem.name = "bob";
 
    setUpElem(elem, &newElem);
}

with the function call setUpElem, I am passing the address of the newElem because that is the pointer (the address memory location).

So to start with the recursive functions, lets print out all of the elements,

// just print out the top element 
void printOutElem(Elem *elem)
{
	cout << "Name : " << elem->name << endl;
}
 
// recursive loop to print them all out
void printThemAllOut(Elem *list)
{
	if (list != NULL)
	{
		printOutElem(list);
		printThemAllOut(list->next);
	}
}

printOutElem will print out the element name associated with the parameter, and the cool recursive function called printThemAllOut, which just re-calls itself with the link to the next element to print out.. just a nice little function..

The next recursive that makes allot more sense, is when you are adding a element instead to the head of the list, but in the middle, like it is sorted, then if you was using loops you would require to keep hold of the previous element so that you can insert the new element into that previous elements -> next link and re-link the inserted element -> next to the current element in the list, hopefully the code helps to understand..

void insertIntoMiddle(Elem *&elem, Elem *midElem)
{
	Elem *cur, *pre= NULL;
 
	for (cur=elem; cur != NULL; cur= cur->next)
	{
		if (cur->name > midElem->name) break;
		// store where we was
		pre = cur;
	}
 
	// place the middle next element equal to where we are in the list
	midElem->next = cur;	
	// if the previous is not null, e.g. previous was somewhere in the list
	if (pre != NULL)
		pre->next = midElem;  // update the previous link to the middle element
	else
		elem = midElem;			// else probably is empty !!.. not good.. 
}

and here is a very nice recursive function that will do the same as above but just in recursive calls to itself with passing the previous link to itself so there is no need to keep a element of the previous link, very nice and allot more easier to debug since less code.

void insertIntoMiddleRecursive(Elem *&elem, Elem *midElem)
{
	if (elem == NULL || elem->name > midElem->name)
	{
		midElem->next = elem;
		elem = midElem;
	}
	else
		insertIntoMiddleRecursive(elem->next, midElem);
}

here is the full source code

#include <string>
#include <iostream>
 
using namespace std;
 
// the element structure, the name is the name of the person
// and the next is a link to next name
struct Elem {
	string name;
	Elem *next;
};
 
// inserts a Elem into a linked list
void setUpElem(Elem * & list, Elem *addnew)
{
	addnew->next = list;
	list = addnew;
}
 
// just print out the top element 
void printOutElem(Elem *elem)
{
	cout << "Name : " << elem->name << endl;
}
 
// recursive loop to print them all out
void printThemAllOut(Elem *list)
{
	if (list != NULL)
	{
		printOutElem(list);
		printThemAllOut(list->next);
	}
}
 
// you have to keep in contact with the current and previous 
// (since that is where we need to place the new element
void insertIntoMiddle(Elem *&elem, Elem *midElem)
{
	Elem *cur, *pre= NULL;
 
	for (cur=elem; cur != NULL; cur= cur->next)
	{
		if (cur->name > midElem->name) break;
		// store where we was
		pre = cur;
	}
 
	// place the middle next element equal to where we are in the list
	midElem->next = cur;	
	// if the previous is not null, e.g. previous was somewhere in the list
	if (pre != NULL)
		pre->next = midElem;  // update the previous link to the middle element
	else
		elem = midElem;			// else probably is empty !!.. not good.. 
}
 
// a far better way of doing this is to use recursive for inserting
// since the parameter is the previous link :)
// this is just a far nicer way of doing the above function
void insertIntoMiddleRecursive(Elem *&elem, Elem *midElem)
{
	if (elem == NULL || elem->name > midElem->name)
	{
		midElem->next = elem;
		elem = midElem;
	}
	else
		insertIntoMiddleRecursive(elem->next, midElem);
}
 
int main()
{
	Elem *elem = NULL;
 
	Elem newElem;
	newElem.name = "steve";
	setUpElem(elem, &newElem);
	Elem nextElem;
	nextElem.name = "genux";
	setUpElem(elem, &nextElem);
 
	printThemAllOut(elem);
 
	// now lets insert into the middle of the linked list
	Elem midElem;
	midElem.name = "henry";
	insertIntoMiddle(elem, &midElem);
 
	cout << "after the insert of \"henry\" into the middle" << endl;
	printThemAllOut(elem);
 
	Elem newMidElem;
	newMidElem.name = "janet";
	insertIntoMiddleRecursive(elem, &newMidElem);
 
	cout << "after the insert of \"janet\" into the middle - using recursive :) " << endl;
	printThemAllOut(elem);
 
 
 
    return 0;
}

and here is the output

Name : genux
Name : steve
after the insert of "henry" into the middle
Name : genux
Name : henry
Name : steve
after the insert of "janet" into the middle - using recursive :)
Name : genux
Name : henry
Name : janet
Name : steve

One thought on “Linked list – recursive functions”

  1. Using break with an if statement isn’t ideal programming. That just shows that you don’t know how to terminate a loop properly. That’s something to think about, other than that this helped me a lot, thanks.

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