Python program to find middle of a linked list using one traversal
Given a singly linked list, find middle of the linked list. Given a singly linked list, find middle of the linked list. For example, if given linked list is 1->2->3->4->5 then output should be 3.
Method 1:
Traverse the whole linked list and count the no. of nodes. Now traverse the list again till count/2 and return the node at count/2.
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Method 2:
Traverse linked list using two pointers. Move one pointer by one and other pointer by two. When the fast pointer reaches end slow pointer will reach middle of the linked list.
# Python 3 program to find the middle of a # given linked list # Node class class Node: # Function to initialise the node object def __init__(self, data): self.data = data self.next = None class LinkedList: def __init__(self): self.head = None def push(self, new_data): new_node = Node(new_data) new_node.next = self.head self.head = new_node # Function to get the middle of # the linked list def printMiddle(self): slow_ptr = self.head fast_ptr = self.head if self.head is not None: while (fast_ptr is not None and fast_ptr.next is not None): fast_ptr = fast_ptr.next.next slow_ptr = slow_ptr.next print("The middle element is: ", slow_ptr.data) # Driver codelist1 = LinkedList()list1.push(5)list1.push(4)list1.push(2)list1.push(3)list1.push(1)list1.printMiddle() |
The middle element is: 2
Method 3:
Initialized the temp variable as head
Initialized count to Zero
Take loop till head will become Null(i.e end of the list) and increment the temp node when count is odd only, in this way temp will traverse till mid element and head will traverse all linked list. Print the data of temp.
# Python 3 program to find the middle of a # given linked list class Node: def __init__(self, value): self.data = value self.next = None class LinkedList: def __init__(self): self.head = None # create Node and and make linked list def push(self, new_data): new_node = Node(new_data) new_node.next = self.head self.head = new_node def printMiddle(self): temp = self.head count = 0 while self.head: # only update when count is odd if (count & 1): temp = temp.next self.head = self.head.next # increment count in each iteration count += 1 print(temp.data) # Driver codellist = LinkedList() llist.push(1)llist.push(20) llist.push(100) llist.push(15) llist.push(35)llist.printMiddle()# code has been contributed by - Yogesh Joshi |
100
