# Program to find the middle node of a singly linked list in Python

PythonServer Side ProgrammingProgramming

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Suppose we have a singly linked list node, we have to find the value of the middle node. And when there are two middle nodes, then we will return the second one. We have to try to solve this in single pass.

So, if the input is like [5,9,6,4,8,2,1,4,5,2], then the output will be 2.

To solve this, we will follow these steps−

• p:= node

• d:= 0, l:= 0

• while node is not null, do

• if d is not same as 2, then

• node:= next of node

• l := l + 1, d := d + 1

• otherwise,

• p:= next of p, d:= 0

• return val of p when l is odd otherwise value of next of p

Let us see the following implementation to get better understanding

## Example

class ListNode:
def __init__(self, data, next = None):
self.val = data
self.next = next
def make_list(elements):
for element in elements[1:]:
while ptr.next:
ptr = ptr.next
ptr.next = ListNode(element)
class Solution:
def solve(self, node):
p=node
d=0
l=0
while node:
if d!=2:
node=node.next
l+=1
d+=1
else:
p=p.next
d=0
return p.val if l & 1 else p.next.val
ob = Solution()
print(ob.solve(head))
Input:
[5,9,6,4,8,2,1,4,5,2]
2