Write a function link_to_list that takes in a linked list and returns the sequence as a Python list. You may assume that the input list is shallow; none of the elements is another linked list. Try to find both an iterative and recursive solution for this problem!其中class Link: empty = () def init(self, first, rest=empty): assert rest is Link.empty or isinstance(rest, Link) self.first = first self.rest = rest def repr(self): if self.rest is not Link.empty: rest_repr = ', ' + repr(self.rest) else: rest_repr = '' return 'Link(' + repr(self.first) + rest_repr + ')' def str(self): string = '<' while self.rest is not Link.empty: string += str(self.first) + ' ' self = self.rest return string + str(self.first) + '>'

Sure! Here's an iterative solution for the function link_to_list:

def link_to_list_iter(link):
    result = []
    while link is not Link.empty:
        result.append(link.first)
        link = link.rest
    return result

And here's a recursive solution:

def link_to_list_rec(link):
    if link is Link.empty:
        return []
    else:
        return [link.first] + link_to_list_rec(link.rest)

Both functions take a linked list as input and return it as a Python list. Hope this helps!

相关问题

Write a function link_to_list that takes in a linked list and returns the sequence as a Python list. You may assume that the input list is shallow; none of the elements is another linked list. Try to find both an iterative and recursive solution for this problem!

Iterative solution:

def link_to_list(head):
    lst = []
    current = head
    while current:
        lst.append(current.val)
        current = current.next
    return lst

This solution traverses the linked list from the head node to the last node, appending the value of each node to a Python list. Once the end of the linked list is reached (i.e., the current node is `None`), the list is returned.

Recursive solution:

def link_to_list(head):
    if head is None:
        return []
    else:
        return [head.val] + link_to_list(head.next)

This solution is a recursive implementation of the same logic. The base case is when the head node is `None`, in which case an empty list is returned. Otherwise, the function recursively calls itself on the next node in the linked list and appends the value of the current node to the resulting list. The final result is the concatenation of all these lists.

1. The following code implements a linked-list Node: struct Node { int data; Node *next; }; Node *A; a) Write a new function to delete the Nth element in the linked-list. The new function's prototype is: bool DeleteNthNode(Node * & listpointer, int nth_element); If the deletion is completed, the function should return true. If the linked-list has less than N elements, then the function should return false. Make sure that your code covers all cases: when the list is empty, when the list only has one element, or two elements etc.

Here's an implementation of the DeleteNthNode function:

bool DeleteNthNode(Node *&listpointer, int nth_element) {
    // Check if the list is empty
    if (listpointer == NULL) {
        return false;
    }
    
    // Check if the first node needs to be deleted
    if (nth_element == 1) {
        Node *temp = listpointer;
        listpointer = listpointer->next;
        delete temp;
        return true;
    }
    
    // Traverse the list to find the node before the nth node
    Node *prev = NULL;
    Node *curr = listpointer;
    int count = 1;
    while (count < nth_element && curr != NULL) {
        prev = curr;
        curr = curr->next;
        count++;
    }
    
    // Check if the nth node was found
    if (curr == NULL) {
        return false;
    }
    
    // Delete the nth node
    prev->next = curr->next;
    delete curr;
    return true;
}

This implementation checks for all the cases mentioned in the prompt - empty list, deleting the first node, and deleting a node in the middle of the list. The function takes a reference to the listpointer so that it can modify the list if necessary.