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Middle of the Linked List

Asked at:AmazonGoogle

Given the head of a singly linked list, return the middle node. If the list has an even number of nodes, return the second middle node.

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Problem

Given the head of a singly linked list, return the middle node of the linked list. If there are two middle nodes, return the second middle node.

Input

The head of a singly linked list.

Output

The middle node of the linked list (for even length, the second of the two middle nodes).

Examples

Input: head = [1,2,3,4,5]

Output: [3,4,5]

Middle node is 3. Return that node (and everything after it).

Input: head = [1,2,3,4,5,6]

Output: [4,5,6]

Two middle nodes are 3 and 4. Return the second one (4).

The brute-force approach

Count the total number of nodes in one pass, then traverse to the middle in a second pass.

count = 0
curr = head
while curr:
    count += 1
    curr = curr.next

curr = head
for _ in range(count // 2):
    curr = curr.next

return curr

Two separate passes over the list. Not wrong, but a fast and slow pointer can find the middle in a single pass without needing to count first.

Time: O(n)Space: O(1)

Spotting the pattern

This is a Linked List problem. The key question to ask yourself:

Without knowing the length of the list, how do I find the middle in one pass?

Answering that is where it clicks, and it's exactly what the guided walkthrough below builds with you: the pattern reasoning, a progressive hint ladder that never spoils the answer, a row-by-row dry run, the optimized solution, and an in-browser editor to run your code against real test cases.

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The Linked List walkthrough for Middle of the Linked List: the progressive hint ladder, a row-by-row dry run, the optimized code, and an in-browser runner, plus the full guided learning path and every other premium problem.

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