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How do you find the middle of a linked list?

Q: How do you find the middle of a linked list?

**Intuition.** You can't index a linked list, and you don't know its length up front. The trick is **two pointers at different speeds**: a slow one steps by 1, a fast one by 2. When the fast pointer reaches the end, the slow pointer is exactly halfway. One pass, no length count. slow +1, fast +2 → slow stops at the middle 1 <text x="150" y="75" fill="currentColor"

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linked-listtwo-pointers

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Slow / fast pointers

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class ListNode:
    def __init__(self, val, nxt=None):
        self.val, self.next = val, nxt

def middle_node(head):
    slow = fast = head
    while fast and fast.next:
        slow = slow.next
        fast = fast.next.next
    return slow


# --- demo ---
def build(vals):
    head = None
    for v in reversed(vals): head = ListNode(v, head)
    return head

print(middle_node(build([1, 2, 3, 4, 5])).val)    # 3
print(middle_node(build([1, 2, 3, 4, 5, 6])).val) # 4 (second middle)

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