Problem $3.$ Extend the LinkedList class adding the following new

methods:

a$)[15$ points$]$ The method called first$\_$index$\_$of$($x$)$ that returns the

index of the first from the left node with value x$.$ If there is no such

node, return None.

The index here is how many nodes away from the head our node is$.$

E$.$g$.,$ applying first$\_$index$\_$of$(5)$ to the linked list $[1->7->5->$

$4->2],$ we get $2$; applying it to $[1->2->3->4->5->6->5],$

we get $4.$

The run$-$time must be in O$($n$).$

b$)[15$ points$]$ The method called get$\_$middle$()$ that returns a list of

values of the middle node$($s$)$ of a linked list. If the list has an odd

number of nodes, there will be only one middle node; otherwise,

there will be two middle nodes.

E$.$g$.,$ applying get$\_$middle$()$ to the linked list $[1->7->5->4->2],$

we get $[5]$; applying it to $[1->2->3->4->5->6],$ we get $[3,4].$

The run$-$time must be in O$($n$).$