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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;
    }
};

• Java
• C++
• Python
• Go
• TypeScript
• Php
• RenderScript

• /**
   * 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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