# middle-of-the-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.

Idea is to use slow and fast (turtle and rabbit) two nodes from head, slow node move 1 node each time,

and fast node move 2 nodes each time, until fast is out of node, slow node is in the middle of the list node, return slow.

middle of the list node

**Time complexity**:

O(N)

N - the length of linkedlist

//Definition for singly-linked list.

public class ListNode {

int val;

ListNode next;

ListNode(int x) { val = x; }

}

class Solution {

public ListNode middleNode(ListNode head) {

if (head == null || head.next == null) return head;

ListNode slow = head;

ListNode fast = head;

while (fast != null && fast.next != null) {

slow = slow.next;

fast = fast.next.next;

}

return slow;

}

}

Last modified 2yr ago