Middle of the Linked List
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 currTwo 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.
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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