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−
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
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
class ListNode: def __init__(self, data, next = None): self.val = data self.next = next def make_list(elements): head = ListNode(elements) for element in elements[1:]: ptr = head while ptr.next: ptr = ptr.next ptr.next = ListNode(element) return head 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() head = make_list([5,9,6,4,8,2,1,4,5,2]) print(ob.solve(head))