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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;
    }
};
  • /**
     * 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;
        }
    }
    
    ############
    
    /**
     * Definition for singly-linked list.
     * public class ListNode {
     *     int val;
     *     ListNode next;
     *     ListNode() {}
     *     ListNode(int val) { this.val = val; }
     *     ListNode(int val, ListNode next) { this.val = val; this.next = next; }
     * }
     */
    class Solution {
        public ListNode middleNode(ListNode 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, val=0, next=None):
    #         self.val = val
    #         self.next = next
    class Solution:
        def middleNode(self, head: ListNode) -> ListNode:
            slow = fast = head
            while fast and fast.next:
                slow, fast = slow.next, fast.next.next
            return slow
    
    ############
    3
    # 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
    
  • /**
     * Definition for singly-linked list.
     * type ListNode struct {
     *     Val int
     *     Next *ListNode
     * }
     */
    func middleNode(head *ListNode) *ListNode {
    	slow, fast := head, head
    	for fast != nil && fast.Next != nil {
    		slow, fast = slow.Next, fast.Next.Next
    	}
    	return slow
    }
    
  • /**
     * Definition for singly-linked list.
     * class ListNode {
     *     val: number
     *     next: ListNode | null
     *     constructor(val?: number, next?: ListNode | null) {
     *         this.val = (val===undefined ? 0 : val)
     *         this.next = (next===undefined ? null : next)
     *     }
     * }
     */
    
    function middleNode(head: ListNode | null): ListNode | null {
        let fast = head,
            slow = head;
        while (fast != null && fast.next != null) {
            fast = fast.next.next;
            slow = slow.next;
        }
        return slow;
    }
    
    
  • /**
     * Definition for a singly-linked list.
     * class ListNode {
     *     public $val = 0;
     *     public $next = null;
     *     function __construct($val = 0, $next = null) {
     *         $this->val = $val;
     *         $this->next = $next;
     *     }
     * }
     */
    class Solution {
        /**
         * @param ListNode $head
         * @return ListNode
         */
        function middleNode($head) {
            $count = 0;
            $tmpHead = $head;
            while ($tmpHead != null) {
                $tmpHead = $tmpHead->next;
                $count++;
            }
            $len = $count - floor($count / 2);
            while ($count != $len) {
                $head = $head->next;
                $count--;
            }
            return $head;
        }
    }
    
    
  • // Definition for singly-linked list.
    // #[derive(PartialEq, Eq, Clone, Debug)]
    // pub struct ListNode {
    //   pub val: i32,
    //   pub next: Option<Box<ListNode>>
    // }
    //
    // impl ListNode {
    //   #[inline]
    //   fn new(val: i32) -> Self {
    //     ListNode {
    //       next: None,
    //       val
    //     }
    //   }
    // }
    impl Solution {
        pub fn middle_node(head: Option<Box<ListNode>>) -> Option<Box<ListNode>> {
            let mut slow = &head;
            let mut fast = &head;
            while fast.is_some() && fast.as_ref().unwrap().next.is_some() {
                slow = &slow.as_ref().unwrap().next;
                fast = &fast.as_ref().unwrap().next.as_ref().unwrap().next;
            }
            slow.clone()
        }
    }
    
    

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