c++ linked list

#include using namespace std;

struct Node { int data; Node* next; };

/* Task 1: Implement the following function for adding a new node to the front of the given list with double the given value.

input Nodes* head: a head pointer to a list int val: an integer that represent a value.

return the head pointer of the new list after the insertion

Example: add a node with given value 9 before HEAD->1->2->3->NULL after HEAD->18->1->2->3->NULL

*/

Node* addDouble(Node* head, int val){

fill in here????

}

/* Task 2: Implement the following function for summing up the elements in the even position of the given list. Assume the head node is in position 1.

input Nodes* head: a head pointer to a list

return void output The sum of all elements in the even position of a given list

Example: input HEAD->1->2->3->9->12->NULL output 11 (2+9=11)

*/

void sumEven(Node* head){ fill in here????

}

/* Task 3: Implement the following function for reversing the node with the minimum value between Node n1 and Node n2.

input Nodes* n1: a pointer to a node in a list Nodes* n2: a pointer to a node in the same list

return the node got removed which has the minimum value between the two given nodes.

Example:

before HEAD->1->2->3->9->12->NULL

after HEAD->1->3->2->9->12->NULL

*/

Node* reverse(Node* n1, Node* n2){ fill in here??????

}

// You are not supposed to change the main function int main() {

Node* head = NULL; Node *p4, *p7;

for(int i = 1; i < 10; i++) { head = addDouble(head, i); if (i==4) p4 = head; if (i==7) p7 = head; } // at this point, we have created the following list: HEAD->18->16->14->12->10->8->6->4->2->NULL // p4 now points to node 4 (the node containing 8); p7 now points to node 7 (the node containing 14)

reverse(p4, p7); // between p4 and p7, there are 2 nodes. // The resulting list is HEAD->18->16->14->10->12->8->6->4->2->NULL

//You can uncomment this line to test. //the output should be: 38 //Please remember to comment this line out before submitting // cout <- sumEven(head);

head = addDouble(head, 16); head = addDouble(head, 20); // at this point, the list is: HEAD->40->32->18->16->14->10->12->8->6->4->2->NULL

reverse(p7, head); // between p7 and head, there are 3 nodes (32, 18 and 16). //The list after the function call is: HEAD->40->16->18->32->14->10->12->8->6->4->2->NULL cout << sumEven(head); // the output should be: 70

// free all nodes Node* tmp; while (head) { tmp = head; head = head->next; delete tmp; } return 0; }