CODEWITHSUNDEEP
javalistlinked-listsingly-linked-list
user4142283
user4142283

Reputation:

How to insert and element before another in a linked list

public void insertElementBefore(E element, E newElement) {
    MyNode<E> current = head;
    if (head != null) {
        while (current != null) {
            if (current.data.equals(element)) {
                MyNode<E> n = new MyNode<E>(newElement);
                n.next = current.next;
                current.next = n;
                return;
            }
            current = current.next;
        }
    }
}

This is what I have for this. I'm having troubles to insert the newElement before intended element. Can't seem to figure out the syntax for it. I've been tinkering with it for a while and the best I could get was for it to insert after the element like it currently does

Any help would be greatly appreciated

Upvotes: 0

Views: 7852

Answers (4)

susheelbhargavk
susheelbhargavk

Reputation: 97

Node n=headNode;
    Node prev=null;
    while(n!=null){
        if(n.getData()==node){
            Node newNode=new Node(data);
            prev.setNext(newNode);
            newNode.setNext(n);
            break;
        }
        else{
            prev=n;
            n=n.getNext();
        }
    }
    if(n.getNext()==null){
        Node newNode= new Node(data);
        prev.setNext(newNode);
        newNode.setNext(null);

    }
    System.out.println("New LinkedList after insert before is:");
    printList();
}

Create a new node called previous-node, and keep track of current element. In my code, if the current element matches the value of the node variable, we add a new node before it.

Upvotes: 0

Vidhi
Vidhi

Reputation: 1

void Insert_Before(int num)
{
    Node *x=new Node();
    x->data=num;

    if (head==NULL) {
        x->next=head;
        head=x;
    } else {
        int c=1;
        cout<<"Element before which insertion has to take place:";
        cin>>n;
        Node *temp=head;
        Node *temp1;

        //check whether the element is present or not
        while (temp->data!=n) { //if present
            temp=temp->next;
            c=c+1; //finds the position of the element n
        }

        x->next=temp;

        if (c==1) {
            head=x;
        } else {
            int i=1;
            while (i<=c-1) { 
                temp1=temp1->next;
                i=i+1;
            }
            temp1->next=x; 
        }
    }
} //Insert_Before

Upvotes: 0

Joop Eggen
Joop Eggen

Reputation: 109593

The trick is to memorise the previous node.

    MyNode<E> current = head;
    MyNode<E> previous = null;
    while (current != null && !current.data.equals(element)) {
            return;
        }
        previous = current;
        current = current.next;
    }
    MyNode<E> n = new MyNode<>(newElement);
    n.next = current;
    if (previous == null) {
       head = n;
    } else {
       previous.next = n;
    }

Upvotes: 1

Luiggi Mendoza
Luiggi Mendoza

Reputation: 85789

In case of a single linked list, you will need two temporary nodes:

  • MyNode<E> current that will represent the current node in the single linked list.
  • MyNode<E> prev that will represent a node before the current node in the single linked list.

Then, you have to add the new node between these nodes. If you don't have the prev node, then when setting the current node as the next node of the new node, then all the nodes before current will be lost.

This is how your code would look like:

public void insertElementBefore(E element, E newElement) {
    MyNode<E> current = head;
    //check here
    MyNode<E> prev = null;
    if (head != null) {
        while (current != null) {
            if (current.data.equals(element)) {
                MyNode<E> n = new MyNode<E>(newElement);
                n.next = current;
                //check here
                if (prev != null) {
                    prev.next = n;
                }
                return;
            }
            //check here
            prev = current;
            current = current.next;
        }
    }
}

Upvotes: 2

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