Python - Delete odd nodes of the Circular Singly Linked List
Deleting odd nodes of a circular singly linked list requires traverse through the list and deleting odd nodes one by one. It involves the following process. If the list contains only head then make the head null. If the list contains more than one element then delete the head and adjust the link of last element with the new head. If next of head is not head then, create three nodes - evenNode pointing to head, oddNode pointing to next of head and temp to store last even node. Then delete oddNode and update evenNode and oddNode to next set of odd-even nodes and continue the process till any of the node reaches head. After that if evenNode reaches head, make next of temp as head else make next of evenNode as head.
The function deleteOddNodes is created for this purpose. It is a 5-step process.
def deleteOddNodes(self): #1. if head is the only element element in # list make the head as null if(self.head != None and self.head.next == self.head): self.head = None elif (self.head != None): #2. if the list contains more than one element # delete the head and adjust the link of # last element with the new head temp = self.head while(temp.next != self.head): temp = temp.next temp.next = self.head.next self.head = None self.head = temp.next #3. create evenNode node - pointing to head # oddNode node - pointing to next of head # temp node - to store last even node if(self.head != None and self.head.next != self.head): evenNode = self.head oddNode = self.head.next while(True): #4. delete odd node and update evenNode and # oddNode to next set of odd-even nodes update # temp node to latest evenNode node continue # the process till any of the node reaches head temp = evenNode evenNode.next = oddNode.next oddNode = None evenNode = evenNode.next oddNode = evenNode.next if(evenNode == self.head or oddNode == self.head): break #5. if evenNode reaches head, make next of temp # as head else make next of evenNode as head if(evenNode == self.head): temp.next = self.head else: evenNode.next = self.head
The below is a complete program that uses above discussed concept of deleting odd nodes of a circular singly linked list.
# node structure class Node: def __init__(self, data): self.data = data self.next = None #class Linked List class LinkedList: def __init__(self): self.head = None #Add new element at the end of the list def push_back(self, newElement): newNode = Node(newElement) if(self.head == None): self.head = newNode newNode.next = self.head return else: temp = self.head while(temp.next != self.head): temp = temp.next temp.next = newNode newNode.next = self.head #delete odd nodes of the list def deleteOddNodes(self): if(self.head != None and self.head.next == self.head): self.head = None elif (self.head != None): temp = self.head while(temp.next != self.head): temp = temp.next temp.next = self.head.next self.head = None self.head = temp.next if(self.head != None and self.head.next != self.head): evenNode = self.head oddNode = self.head.next while(True): temp = evenNode evenNode.next = oddNode.next oddNode = None evenNode = evenNode.next oddNode = evenNode.next if(evenNode == self.head or oddNode == self.head): break if(evenNode == self.head): temp.next = self.head else: evenNode.next = self.head #display the content of the list def PrintList(self): temp = self.head if(temp != None): print("The list contains:", end=" ") while (True): print(temp.data, end=" ") temp = temp.next if(temp == self.head): break print() else: print("The list is empty.") # test the code MyList = LinkedList() #Add five elements in the list. MyList.push_back(10) MyList.push_back(20) MyList.push_back(30) MyList.push_back(40) MyList.push_back(50) #Display the content of the list. MyList.PrintList() #delete odd nodes of the list MyList.deleteOddNodes() print("After deleting odd nodes:") #Display the content of the list. MyList.PrintList()
The above code will give the following output:
The list contains: 10 20 30 40 50 After deleting odd nodes: The list contains: 20 40