Delete nodes having greater value on right

PROBLEM :

Given a singly linked list, remove all the nodes which have a greater value on right side.

Input:
You have to complete the method which takes 1 argument: the head of the  linked list  .You should not read any input from stdin/console. There are multiple test cases. For each test case, this method will be called individually.

Output:
Your function should return a pointer to the linked list in which all nodes which have a greater value on right side are removed.

Constraints:
1<=T<=50
1<=size of linked list <=100
1<=element of linked list <=1000

Example:

Input:
3
8
12 15 10 11 5 6 2 3
6
10 20 30 40 50 60
6
60 50 40 30 20 10

Output:

15 11 6 3
60
60 50 40 30 20 10

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SIMPLE c++ IMPLEMENTATION :
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/*

The structure of linked list is the following

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

*/

struct Node* reverse(struct Node *) ;

Node *compute(Node *head)
{
    head=reverse(head) ;
    struct Node *temp,*ptr ;
    temp=head ;
    int max=temp->data ;  
    while(temp->next!=NULL)
    {
        if(temp->next->data<max)
        {
            ptr=temp->next ;
            temp->next=ptr->next ;
            free(ptr) ;
        }
        else if(temp->next->data>max)
            max=temp->next->data ;
        else
            temp=temp->next ;
    }
    head=reverse(head) ;
    return head ;
}

struct Node* reverse(struct Node *head)
{
    if(head==NULL)
    return head ;
   
    if(head->next==NULL)
    return head;
   
    struct Node *one,*two,*third ;
    one=head ;
    two=one->next ;
    third=two->next ;
   
    one->next=NULL ;
    two->next=one ;
   
    while(third!=NULL)
    {
        one=two ;
        two=third ;
        third=third->next ;
       
        two->next=one ;
    }
    return two ;
   
}

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