Solution Approach

The implementation uses the fast and slow pointer technique (Floyd's Tortoise and Hare algorithm) to find the middle node in a single pass.

Algorithm Steps:

1. Initialize two pointers: Both slow and fast start at the head of the linked list

   slow = fast = head

2. Traverse the linked list with different speeds:

   • The loop continues while both fast and fast.next exist
   • This condition ensures we don't get a null pointer error when fast tries to jump two nodes

   while fast and fast.next:

3. Move the pointers:

   • slow moves one step forward: slow = slow.next
   • fast moves two steps forward: fast = fast.next.next
   • Python allows simultaneous assignment, so we can write:

   slow, fast = slow.next, fast.next.next

4. Return the middle node: When the loop exits, slow points to the middle node

   return slow

Why the loop condition works:

• fast checks if the fast pointer itself exists (prevents error when list has odd length)
• fast.next checks if fast can make another two-step jump (prevents error when list has even length)
• When either becomes None, we've reached or passed the end of the list

Example walkthrough:

• For list 1->2->3->4->5:

  • Initially: slow at 1, fast at 1
  • Iteration 1: slow at 2, fast at 3
  • Iteration 2: slow at 3, fast at 5
  • Loop ends (fast.next is None), return node 3

• For list 1->2->3->4:

  • Initially: slow at 1, fast at 1
  • Iteration 1: slow at 2, fast at 3
  • Iteration 2: slow at 3, fast is None
  • Loop ends, return node 3 (the second middle node)

Time Complexity: O(n) where n is the number of nodes - we traverse at most n/2 iterations Space Complexity: O(1) - only using two pointers regardless of input size

Example Walkthrough

Let's walk through finding the middle node of the linked list: 1 -> 2 -> 3 -> 4 -> 5

Initial Setup:

• Both slow and fast pointers start at the head (node 1)
• List visualization: [1] -> 2 -> 3 -> 4 -> 5
  • slow points to node 1
  • fast points to node 1

Iteration 1:

• Check condition: fast (node 1) exists ✓ and fast.next (node 2) exists ✓
• Move pointers:
  • slow moves one step: slow = slow.next → now at node 2
  • fast moves two steps: fast = fast.next.next → now at node 3
• List visualization: 1 -> [2] -> (3) -> 4 -> 5
  • [2] = slow pointer position
  • (3) = fast pointer position

Iteration 2:

• Check condition: fast (node 3) exists ✓ and fast.next (node 4) exists ✓
• Move pointers:
  • slow moves one step: now at node 3
  • fast moves two steps: now at node 5
• List visualization: 1 -> 2 -> [3] -> 4 -> (5)
  • [3] = slow pointer position
  • (5) = fast pointer position

Iteration 3:

• Check condition: fast (node 5) exists ✓ but fast.next is None ✗
• Loop terminates
• Return slow which points to node 3

Result: Node 3 is correctly identified as the middle node of the 5-node list.

For an even-length list like 1 -> 2 -> 3 -> 4, the algorithm would proceed:

• Start: slow and fast at node 1
• After iteration 1: slow at node 2, fast at node 3
• After iteration 2: slow at node 3, fast becomes None (moved past node 4)
• Returns node 3, which is the second middle node (nodes 2 and 3 are both "middle", we return the second one)

Time and Space Complexity

Time Complexity: O(n) where n is the number of nodes in the linked list. The algorithm uses two pointers - a slow pointer that moves one step at a time and a fast pointer that moves two steps at a time. The fast pointer will reach the end of the list after traversing approximately n/2 iterations (since it moves twice as fast), at which point the loop terminates. Therefore, the slow pointer visits approximately n/2 nodes and the fast pointer visits approximately n nodes, resulting in a total time complexity of O(n).

Space Complexity: O(1) - The algorithm only uses two pointer variables (slow and fast) regardless of the size of the input linked list. No additional data structures or recursive calls are made that would scale with the input size, making the space complexity constant.

Learn more about how to find time and space complexity quickly.

Common Pitfalls

1. Incorrect Loop Condition

A common mistake is using only while fast.next: or while fast: as the loop condition instead of while fast and fast.next:.

Why it's wrong:

• Using only while fast.next: will cause a NullPointerException/AttributeError when fast itself becomes None (happens with odd-length lists)
• Using only while fast: will cause an error when trying to access fast.next.next if fast.next is None (happens with even-length lists)

Incorrect implementation:

# This will fail for odd-length lists while fast.next: # Error when fast is None  slow = slow.next  fast = fast.next.next

Correct implementation:

# Check both fast and fast.next while fast and fast.next:  slow = slow.next  fast = fast.next.next

2. Moving Pointers in Wrong Order

Another pitfall is updating the pointers in the wrong sequence when not using simultaneous assignment.

Why it's wrong: If you update fast first using sequential assignments, you lose the reference needed for updating slow.

Incorrect implementation:

while fast and fast.next:  fast = fast.next.next # Wrong: updating fast first  slow = slow.next # This still works but can lead to confusion

Better implementations:

# Option 1: Update slow first while fast and fast.next:  slow = slow.next  fast = fast.next.next  # Option 2: Use simultaneous assignment (Python feature) while fast and fast.next:  slow, fast = slow.next, fast.next.next

3. Off-by-One Error for Even-Length Lists

Some might mistakenly think they need special handling for even-length lists to return the second middle node.

Why it's wrong: The algorithm naturally returns the second middle node for even-length lists. Adding extra logic is unnecessary and can introduce bugs.

Incorrect approach:

# Unnecessary complexity def middleNode(self, head):  # Count nodes first  count = 0  current = head  while current:  count += 1  current = current.next   # Find middle based on count  middle_index = count // 2  current = head  for _ in range(middle_index):  current = current.next  return current

Correct approach: The two-pointer technique automatically handles both odd and even cases without special logic.

4. Not Handling Edge Cases

While the algorithm handles most cases well, ensure you consider:

• Single node list (works correctly as-is)
• Two node list (returns the second node correctly)

The beauty of the while fast and fast.next: condition is that it handles all these edge cases without additional checks.