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Linked List Middle

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Your question is Linked List Middle. Start with the requirements on the right.

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Problem

A Check Point Security Management component stores an ordered sequence of validation records in a singly linked list. Given the list head, return the value of its middle node efficiently.

Use the fast and slow pointer technique. The slow pointer advances one node at a time, while the fast pointer advances two nodes at a time. If the list contains an even number of nodes, return the second middle node.

Formal Specification

The input is head, either None or a reference to a singly linked ListNode object with fields value and next. Return the integer value stored in the middle node. The test cases represent linked lists as arrays of values, and the test harness converts each array into a linked list before calling the function.

Implement:

def find_middle(head):

Constraints

  • 0 <= n <= 10^5
  • -10^9 <= node.value <= 10^9
  • The linked list is singly linked and acyclic
  • For even n, return the second middle node

Function Signature

def find_middle(head):
Your solutionPython 3
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