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Find a path from A to E
E
B
border
D
explored nodes
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Complete public class Solution { public static LinkedListNode<Integer> findPrevNode(LinkedListNode<Integer> head, int count) { for (int i=0;i<count-1;i++) { head=head.next; } return head; } public static LinkedListNode<Integer> swapNodes(LinkedListNode<Integer> head, int i, int j) { //Your code goes here if (head==null) { return head; } else if (j==0 || (i-j==-1)) { int temp=i; i=j; j=temp; } LinkedListNode<Integer> swap1=null,swap2=null,p1=null,n1=null,p2=null,n2=null; if (i==0 && i-j==1) { swap1=head; swap2=head.next; swap1=swap1.next; head=swap2; swap2.next=swap1; } else if(i-j==1) { p1=findPrevNode(head,j); swap1=p1.next; swap2=swap1.next; n2=swap2.next; //System.out.println(p1.data); //System.out.println(swap1.data);…
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In python. Write a LinkedList class that has recursive implementations of the add and remove methods. It should also have recursive implementations of the contains, insert, and reverse methods. The reverse method should not change the data value each node holds - it must rearrange the order of the nodes in the linked list (by changing the next value each node holds). It should have a recursive method named to_plain_list that takes no parameters (unless they have default arguments) and returns a regular Python list that has the same values (from the data attribute of the Node objects), in the same order, as the current state of the linked list. The head data member of the LinkedList class must be private and have a get method defined (named get_head). It should return the first Node in the list (not the value inside it). As in the iterative LinkedList in the exploration, the data members of the Node class don't have to be private. The reason for that is because Node is a trivial class…
Given a circular linked list in which the last node is connected to the first node. If we set a node curr equal to first node at the start of traversal and we start traversing the circular linked list, the condition to check if we have reached the end of the list then becomes curr = first curr.next == null curr.next.next == first curr.next == first None of the Above
Add more methods to the singly linked list class then test them• search(e) // Return one node with 3 values (stuID, stuName, stuScore) which matches agiven key e (studentID).• addAfter(e, stuID, stuName, stuScore) //Add a new node with 3 values (stuID,stuName, stuScore) after the node with the key e (studentID).• removeAt(e) //Remove a node which matches a given key e (studentID)• count() //Return a number of nodes of list.• update(stuID, stuName, stuScore) //Update the values of one node three codes below two in pictures one,typed out. public class SlinkedList<A,B,C> { private Node head; private Node tail; private int size; public SlinkedList(){ head=null; tail=null; size=0; } public int getSize(){ return size; } public boolean isEmpty(){ return size == 0; } public A getFirstStuId(){ if(isEmpty()) return null; return (A) head.getStuID();}public B getFirstStuName(){ if(isEmpty())…
We can iterate over a hash map using a for-each loop directly. (Exactly like we can over a LinkedList or an ArrayList). True or False
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In a circular linked list, how can you determine if a given node is the last node in the list? Describe the approach you would take to solve this problem.
There are an undetermined number of nodes in the fastest linkedlist.Is there one that stands out as being the longest?
Assume that the nodes of the singly linked lists are arranged in decreasing order of the exponents of the variable x in order to add the two polynomials.The objective is to create a fresh list of nodes that represents the addition of P1 and P2. This is done by adding the COEFF fields of nodes in lists P1 and P2 that have identical powers of variable x, and then making a new node in the resulting list P1 + P2. The key part of the technique is shown below.The start pointers of the singly linked lists that correspond to the polynomials P1 and P2 are P1 and P2, respectively. Two temporary pointers, PTR1 and PTR2, are created with starting values of P1 and P2, respectively. Make procedural code.
bool DelOddCopEven(Node* headPtr) { if (headPtr == nullptr) { return false; } Node* prev = nullptr; Node* current = headPtr; while (current != nullptr) { if (current->data % 2 != 0) { // Odd-valued node, delete it if (prev != nullptr) { prev->link = current->link; } else { headPtr = current->link; } Node* temp = current; current = current->link; delete temp; } else { // Even-valued node, create a copy and insert it after the original node Node* newNode = new Node; newNode->data = current->data; newNode->link = current->link; current->link = newNode; // Update the previous and current pointers prev = newNode; current = newNode->link; } } return true;} int FindListLength(Node* headPtr){ int length = 0;…
You should create a method and call it findTotalleaves (); this method will find all of the nodes that are considered to be leaves, which are nodes that do not have any BNode dependents. The public function must not take any arguments, but the private method must accept as a parameter a BNode that represents the current root.

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