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
    

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