GoLang: Compute the Middle of a Linked List

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

Example 1:
Input: [1,2,3,4,5]
Output: Node 3 from this list (Serialization: [3,4,5])
The returned node has value 3. (The judge’s serialization of this node is [3,4,5]).
Note that we returned a ListNode object ans, such that:
ans.val = 3, ans.next.val = 4, ans.next.next.val = 5, and ans.next.next.next = NULL.

Example 2:
Input: [1,2,3,4,5,6]
Output: Node 4 from this list (Serialization: [4,5,6])
Since the list has two middle nodes with values 3 and 4, we return the second one.

Note:
The number of nodes in the given list will be between 1 and 100.

Middle of the Linked List via GoLang

Two Pointers – fast and slow. The fast pointer moves two steps at a time and the slow pointer moves one step instead. When the fast pointer reaches the end, the slow pointer is in the middle. The time complexity is O(N) where N is the number of the nodes in the linked list. Below is the GoLang implementation of finding the middle of the linked list.

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/**
 * Definition for singly-linked list.
 * type ListNode struct {
 *     Val int
 *     Next *ListNode
 * }
 */
func middleNode(head *ListNode) *ListNode {
    if head == nil || head.Next == nil {
        return head
    }
    var fast, slow = head, head
    for fast != nil && fast.Next != nil {
        fast = fast.Next.Next
        slow = slow.Next
    }
    return slow
}

/**
 * Definition for singly-linked list.
 * type ListNode struct {
 *     Val int
 *     Next *ListNode
 * }
 */
func middleNode(head *ListNode) *ListNode {
    if head == nil || head.Next == nil {
        return head
    }
    var fast, slow = head, head
    for fast != nil && fast.Next != nil {
        fast = fast.Next.Next
        slow = slow.Next
    }
    return slow
}

A shorter implementation:

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/**
 * Definition for singly-linked list.
 * type ListNode struct {
 *     Val int
 *     Next *ListNode
 * }
 */
func middleNode(head *ListNode) *ListNode {
    var mid = head
    for head != nil && head.Next != nil {
        head, mid = head.Next.Next, mid.Next
    }
    return mid
}

/**
 * Definition for singly-linked list.
 * type ListNode struct {
 *     Val int
 *     Next *ListNode
 * }
 */
func middleNode(head *ListNode) *ListNode {
    var mid = head
    for head != nil && head.Next != nil {
        head, mid = head.Next.Next, mid.Next
    }
    return mid
}

See other implementations of getting the middle of the linked list:

• Fast and Slow Pointer to Get the Middle of the Linked List
• Teaching Kids Programming – Fast and Slow Pointer to Obtain the Middle of the Linked List
• How to Compute the Middle of the Linked List using Fast and Slow Pointer?
• GoLang: Compute the Middle of a Linked List

–EOF (The Ultimate Computing & Technology Blog) —