An approach to remove nodes from a linked list that have the same key is provided.

/ This method takes as parameters a reference to the head of a linked list, a    // position specified by n, and a key. It returns the number of occurrences    // of the key in the linked list beginning at the n-th node. If n = 0, it means    // you should search in the entire linked list. If n = 1, then you should skip    // the first node in the list.    public int numOccurrencesRec(LNode node, int n, int key)    {        // TODO: implement this method             return 0; // replace this statement with your own return    }    } LNode Below: public class LNode{    // instance variables    private int m_info;    private LNode m_link;       // constructor    public LNode(int info)    {        m_info = info;        m_link = null;    }       // member methods    public void setLink(LNode link)    {        m_link = link;    }       public LNode getLink()    {        return m_link;    }       public int getInfo()    {        return m_info;    }  }

What is the best number of nodes for a linked list? What is the longest time you have to wait?

python language   Write the lines of code that will print out all of the data in a Linked List.