Reverse a doubly linked list| Linked List
TASK(easy)
Given the pointer to the head node of a doubly linked list, reverse the order of the nodes in place. That is, change the next and prev pointers of the nodes so that the direction of the list is reversed. Return a reference to the head node of the reversed list.
Note: The head node might be NULL to indicate that the list is empty.
Function Description
Complete the reverse function in the editor below.
reverse has the following parameter(s):
- DoublyLinkedListNode head: a reference to the head of a DoublyLinkedList
Returns
- DoublyLinkedListNode: a reference to the head of the reversed list
Input Format
The first line contains an integer t, the number of test cases.
Each test case is of the following format:
- The first line contains an integer n, the number of elements in the linked list.
- The next n lines contain an integer each denoting an element of the linked list.
Constraints
1<=t<=10
1<=n<=1000
1<=DoublyLinkedListNode.data<=1000
Output Format
Return a reference to the head of your reversed list. The provided code will print the reverse array as a one line of space-separated integers for each test case.
Sample Input
1
4
1
2
3
4
Sample Output
4 3 2 1
SOLUTION 1
def reverse(head):
if head.next is None:
head.next = head.prev
head.prev = None
return head
else:
old_next = head.next
head.next = head.prev
head.prev = old_next
return reverse(head.prev)
SOLUTION 2
def reverse(llist):
while llist:
before = llist
llist.prev, llist.next = llist.next, llist.prev
llist = llist.prev
return before
