Deletion and Reverse in Circular Linked List | GeekForGeeks | Problem of the Day

[@mathematics3398]

Table of Contents
0:00 Problem Statement
0:35 Solution - Deletion
4:25 Solution - Reverse
7:30 Pseudo Code - Deletion
12:47 Pseudo Code - Reverse
14:59 Code - Python
17:19 Code - C++

[@mathematics3398]

class Solution {
public:
// Function to reverse a circular linked list
Node* reverse(Node* head) {
if (!head)
return head;
if (head->next == head)
return head;
Node *prvs = NULL;
Node *last = head;
while (head->next != last) {
Node *NEXT = head->next;
head->next = prvs;
prvs = head;
head = NEXT;
}
head->next = prvs;
prvs = head;
last->next = prvs;
return prvs;
}
// Function to delete a node from the circular linked list
Node* deleteNode(Node* head, int key) {
if (!head or head->next == head)
return NULL;
Node *ptr2 = head;
Node *prvs = NULL;
Node *ptr = head;
while(ptr->data != key and ptr->next != head) {
prvs = ptr;
ptr = ptr->next;
}
// return if key not found
if (ptr->data != key)
return head;
if (!prvs) {
while (ptr2->next != head) {
prvs = ptr2;
ptr2 = ptr2->next;
}
prvs = ptr2;
}
prvs->next = ptr->next;
if (ptr == head) {
return ptr->next;
}
return head;
}
};

[@mathematics3398]

class Solution:
# Function to reverse a circular linked list
def reverse(self, head):
if not head or head.next == head:
return head
prvs = None
last = head
while head.next != last:
NEXT = head.next
head.next = prvs
prvs = head
head = NEXT
head.next = prvs
prvs = head
last.next = prvs
return prvs
# Function to delete a node from the circular linked list
def deleteNode(self, head, key):
if not head or head.next == head:
return None
ptr1 = ptr2 = head
prvs = None
while ptr1.data != key and ptr1.next != head:
prvs = ptr1
ptr1 = ptr1.next
if ptr1.data != key:
return head
if not prvs:
while ptr2.next != head:
ptr2 = ptr2.next
prvs = ptr2
prvs.next = ptr1.next
if ptr1 == head:
return ptr1.next
return head