Formatted question description: https://leetcode.ca/all/876.html

876. Middle of the Linked List (Easy)

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.

Solution 1. Two Pointers

// OJ: https://leetcode.com/problems/middle-of-the-linked-list/
// Time: O(N)
// Space: O(1)
class Solution {
public:
    ListNode* middleNode(ListNode* head) {
        auto slow = head, fast = head;
        while (fast && fast->next) {
            slow = slow->next;
            fast = fast->next->next;
        }
        return slow;
    }
};

Java

/**
 * 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, fast = head;
        while (fast != null && fast.next != null) {
            slow = slow.next;
            fast = fast.next.next;
        }
        return slow;
    }
}

// OJ: https://leetcode.com/problems/middle-of-the-linked-list/
// Time: O(N)
// Space: O(1)
class Solution {
public:
    ListNode* middleNode(ListNode* head) {
        auto slow = head, fast = head;
        while (fast && fast->next) {
            slow = slow->next;
            fast = fast->next->next;
        }
        return slow;
    }
};

# Definition for singly-linked list.
# class ListNode:
#     def __init__(self, x):
#         self.val = x
#         self.next = None

class Solution:
    def middleNode(self, head):
        """
        :type head: ListNode
        :rtype: ListNode
        """
        dummy = ListNode(0)
        dummy.next = head
        slow, fast = dummy, dummy
        while fast and fast.next:
            slow = slow.next
            fast = fast.next.next
        return slow.next if fast else slow

876 - Middle of the Linked List

876. Middle of the Linked List (Easy)

Solution 1. Two Pointers

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