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Assoli18 [71]
3 years ago
13

What did Peter and Rosemary do for a living?​

Computers and Technology
1 answer:
marta [7]3 years ago
6 0

Answer:

Peter and Rosemary Grant are distinguished for their remarkable long-term studies demonstrating evolution in action in Galápagos finches. They have demonstrated how very rapid changes in body and beak size in response to changes in the food supply are driven by natural selection.

Explanation:

please

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Answer:

public class CircularList

{

  private ListNode head; // front of the LinkedList

  private ListNode tail; // last node of the LinkedList

  private int size; // size of the LinkedList

   

  // constructs a new CircularList

  public CircularList()

  {

    head = tail = null;

    size = 0;

  }

   

  // returns the size of the array

  public int size()

  {

     return size;

  }

   

  // returns whether the list is empty

  public boolean isEmpty()

  {

     return (size == 0);

  }

   

  // returns the value of the first node

  public Integer first()

  {

     if (head != null) {

       return head.getValue();

     }

     return -1;

  }

   

  // returns the value of the last node

  public Integer last()

  {

     if (tail != null) {

       return tail.getValue();

     }

     return -1;

  }

  // adds a node to the front of the list

  public void addFirst(Integer value)

  {

    head = new ListNode(value, head);

    if (tail == null) {

      tail = head;

    }

    size++;

  }

   

  // adds a node to the end of the list

  public void addLast(Integer value)

  {

    ListNode newTail = new ListNode(value, null);

    if (tail != null) {

      tail.setNext(newTail);

      tail = newTail;

    } else {

      head = tail = newTail;

    }

     

    size++;

  }

   

  // adds a node at the position pos  

  public void addAtPos(int pos, Integer value)

  {

     if (pos == 0) { // Add at the start

       addFirst(value);

       return;

     }

     if (pos <= 0 || pos > size) { // Ignore attempts to add beyond the ends

       return;

     }

     if (pos == size) { // Special case, tail has to be adjusted

       addLast(value);

       return;

     }

     // size and pos are guaranteed both non-zero

     ListNode ptr = head; // ptr is the node before the new one

     for(int i=0; i<pos-1; i++) {

       ptr = ptr.getNext();

     }

     ListNode newNode = new ListNode(value, ptr.getNext());

     ptr.setNext(newNode);

     size++;

  }

   

  // removes the first node and returns the value of the removed node or -1 if the list is empty

  public Integer removeFirst()

  {

     Integer retVal = -1;

     if (head != null) {

       retVal = head.getValue();

       head = head.getNext();

       size--;

     }

     if (size == 0) {

       head = tail = null;

     }

     return retVal;

  }

   

  // removes the node at position pos and returns the value of the removed node or -1 if pos is not a valid position

  public Integer removeNode(int pos)

  {

     Integer retVal = -1;

     if (head == null || pos < 0 || pos >= size) {

       return retVal;

     }

     if (pos == 0) {

       return removeFirst();

     }

     ListNode ptr = head; // ptr is the node before the deleted

     for(int i=0; i<pos-1; i++) {

       ptr = ptr.getNext();

     }

     retVal = ptr.getNext().getValue();

     if (pos == size-1) { // Is it the last element?      

       tail = ptr;

       tail.setNext(null);

     } else {

       ptr.setNext(ptr.getNext().getNext());

     }

     

     size--;

     return retVal;

  }  

   

  // finds and returns the position of find, or -1 if not found

  public int findNode(Integer find)

  {

     ListNode ptr = head;

     for(int pos=0; pos<size; pos++) {

       if (ptr.getValue() == find) {

         return pos;

       }

       ptr = ptr.getNext();

     }

     return -1;

  }  

   

  // rotates the list by placing the first element at the end

  public void rotate()

  {

     addLast(removeFirst());

  }

   

  // returns the list of values in the LinkedList

  public String toString()

  {

     String output = "";

     ListNode iter = head;

     while(iter != null) {

       output += String.format("%d ", iter.getValue());

       iter = iter.getNext();

     }

     return output;

  }

         

}

Explanation:

Enjoy. Linked list are always more complex than you expect. It is a good exercise to try once, then start using libraries. Life is too short to debug linked lists!

7 0
3 years ago
Using the framework below add code to allow the loop to sum the odd numbers from 1 to 10 in the loop and print the sum out at th
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Answer:

Below

Explanation:

#include <stdio.h>

int main(void){

int sum = 0;

for(int i = 1; i <=10; i++){

   if (i % 2 != 0 ){

       sum += i;

   }

   }

printf("%d\n", sum);

return 0;

}//end main

8 0
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